datavorous/entrance-exam-dataset huggingface dataset:
The standard deviation of $a, a+d, a+2 d, \ldots, a+2 n d$ is

Options:
A $n d$
B $n^2 d$
C $\sqrt{\frac{n(n+1)}{3}} d$
D $\sqrt{\frac{n(n+3)}{3}} d$

The correct answer is:
$\sqrt{\frac{n(n+1)}{3}} d$

Solution:
Given series,
$$
a, a+d, a+2 d, \ldots, a+2 n d
$$
Here, $\quad a+2 n d=a+(m-1) d$
$$
\begin{array}{rlrl}
\Rightarrow & & 2 n d=(m-1) d \\
\Rightarrow & 2 n=m-1 \\
\Rightarrow & & 2 n+1=m
\end{array}
$$
$$
\begin{aligned}
& \therefore \text { Standard deviation }=\sqrt{\frac{(2 n+1)^2-1}{12} d^2} \\
&=\sqrt{\frac{(2 n+1-1)(2 n+1+1)}{12} d^2} \\
&=\sqrt{\frac{(2 n+2) 2 n}{12} d^2}=\sqrt{\frac{(n+1) n}{3} d}
\end{aligned}
$$

Topic: Mathematics, Statistics, TS EAMCET, TS EAMCET 2016
The standard deviation of a data is of any change in origin, but is on the change of scale.The sum of the squares of the deviations of the values of the variable is when taken about their arithmentic mean.

Options:

None

Solution:
Independent

Topic: Mathematics, Statistics
The standard deviation of first 10 multiples of 4 is

Options:
A $7$
B $8$
C $11.5$
D $14$

The correct answer is:
$11.5$

Solution:
Entries are $4,8,12,16,20,24,28,32,36,40$.
$$
\begin{aligned}
\text { Variance } & =\frac{\sum_{i=1}^{10}(4 i)^2}{10}-\left(\frac{\sum_{i=1}^{10} 4 i}{10}\right)^2 \\
& =\frac{16 \times 10 \times 11 \times 21}{60}-\left(\frac{4 \times 55}{10}\right)^2 \\
& =132
\end{aligned}
$$
$\therefore$ Standard deviation $=\sqrt{132} \approx 11.5$

Topic: Mathematics, Statistics, AP EAMCET, AP EAMCET 2022 (07 Jul Shift 2)
The standard deviation of $\mathrm{n}$ observations $\mathrm{x}_{1}, \mathrm{x}_{2}, \ldots \ldots \mathrm{x}_{\mathrm{n}}$ is $6 .$ The standard deviation of another set of $\mathrm{n}$ observations $\mathrm{y}_{1}$, $\mathrm{y}_{2}$
$\mathrm{y}_{\mathrm{n}}$ is $8 .$ What is the standard deviation of $\mathrm{n}$ observations $\mathrm{x}_{1}-\mathrm{y}_{1}, \mathrm{x}_{2}-\mathrm{y}_{2}, \ldots \ldots . ., \mathrm{x}_{\mathrm{n}}-\mathrm{y}_{\mathrm{n}} ?$

Options:
A 10
B 7
C 14
D 2

The correct answer is:
2

Solution:
The standard deviation of n observation $\mathrm{x}_{1}, \mathrm{x}_{2}$, $\mathrm{x}_{\mathrm{n}}$ is 6 and of $\mathrm{y}_{1}, \mathrm{y}_{2}, \ldots \ldots, \mathrm{y}_{n}$ is 8, then the standard
deviation of n observation $\mathrm{x}_{1}-\mathrm{y}_{1}, \mathrm{x}_{2}-\mathrm{y}_{2}, \mathrm{x}_{3}-\mathrm{y}_{3}, \ldots \ldots$
$x_{n}-y_{n}$ is $8-6=2$.

Topic: Mathematics, Statistics, NDA, NDA 2006 (Phase 1)
The standard deviation of some consecutive integers is found to be $2 .$ Which of the following statements best describes the nature of the consecutive integers?

Options:
A The integers are any set of eight consecutive integers
B The integers are any set of eight consecutive positive integers
C The integers are any set of seven consecutive integers
D None of the above

The correct answer is:
The integers are any set of seven consecutive integers

Solution:
Since, the standard deviation of same consecutive integers is 2, these integers are any set of seven consecutive integers.

Topic: Mathematics, Statistics, NDA, NDA 2010 (Phase 2)
The standard deviation of the data $6,7,8,9$, 10 is

Options:
A $\sqrt{2}$
B $\sqrt{10}$
C 2
D 10

The correct answer is:
$\sqrt{2}$

Solution:
Given data $6,7,8,9,10$
$\begin{aligned}
\bar{x} &=\frac{6+7+8+9+10}{5}=8 \\
\mathrm{SD} &=\sqrt{\frac{\sum x_{i}^{2}}{n}-(\bar{x})^{2}} \\
\mathrm{SD} &=\sqrt{\frac{6^{2}+7^{2}+8^{2}+9^{2}+10^{2}}{5}-(8)^{2}} \\
\mathrm{SD} &=\sqrt{\frac{36+49+64+81+100-320}{5}} \\
\mathrm{SD} &=\sqrt{\frac{330-320}{5}}=\sqrt{\frac{10}{5}}=\sqrt{2}
\end{aligned}$

Topic: Mathematics, Functions, KCET, KCET 2020
The standard deviation $\sigma$ of the first N natural numbers can be obtained using which one of the following formulae:

Options:
A $\sigma=\frac{\mathrm{N}^{2}-1}{12} \quad$
B $=\sqrt{\frac{N^{2}-1}{12}}$
C $\sigma=\sqrt{\frac{\mathrm{N}-1}{12}}$
D $\sigma=\sqrt{\frac{N^{2}-1}{6 N}}$

The correct answer is:
$=\sqrt{\frac{N^{2}-1}{12}}$

Solution:
$\sigma^{2}=\frac{1}{\mathrm{n}} \sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{x}_{\mathrm{i}}^{2}-\left(\frac{1}{\mathrm{n}} \cdot \sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{x}_{\mathrm{i}}\right)^{2}$
$=\frac{1}{n}\left(1^{2}+2^{2}+3^{2}+\ldots+n^{2}\right)-\left(\frac{1}{n}(1+2+3+\ldots .+n)\right)^{2}$
$=\frac{1}{n} \frac{n(n+1)(2 n+1)}{6}-\left(\frac{1}{n} \cdot \frac{n(n+1)}{2}\right)^{2}$
$=\frac{\mathrm{n}^{2}-1}{12} \quad \therefore \sigma=\sqrt{\frac{\mathrm{n}^{2}-1}{12}} .$

Topic: Mathematics, Statistics, NDA, NDA 2018 (Phase 1)
The standard deviation of the following distribution
\begin{array}{|c|c|c|c|}
\hline C.I. & 0-6 & 6-12 & 12-18 \\
\hline \mathrm{f}_{\mathrm{i}} & 2 & 4 & 6 \\
\hline
\end{array}
is

Options:
A $5 \sqrt{2}$
B $\sqrt {5}$
C $2\sqrt {5}$
D $20$

The correct answer is:
$2\sqrt {5}$

Solution:
\begin{array}{|c|c|c|c|c|c|}
\hline C.I. & \mathbf{f}_{\mathbf{i}} & \boldsymbol{x}_{\mathbf{i}} & \boldsymbol{x}_{\mathbf{i}}^{\mathbf{2}} & \mathbf{f}_{\mathbf{i}} \boldsymbol{x}_{\mathbf{i}} & \mathbf{f}_{\mathbf{i}} \boldsymbol{x}_{\mathbf{i}}^{\mathbf{2}} \\
\hline 0-6 & 2 & 3 & 9 & 6 & 18 \\
\hline 6-12 & 4 & 9 & 81 & 36 & 324 \\
\hline 12-18 & 6 & 15 & 225 & 90 & 1350 \\
\hline Total & \mathbf{1 2} & & & \mathbf{1 3 2} & \mathbf{1 6 9 2} \\
\hline
\end{array}
Here $\sum \mathrm{f}_{\mathrm{i}}=12, \sum \mathrm{f}_{\mathrm{i}} x_{\mathrm{i}}=132, \sum \mathrm{f}_{\mathrm{i}} x_{\mathrm{i}}^2=1692$
$\begin{aligned}
\therefore \quad \mathrm{V}(\mathrm{X}) & =\frac{1692}{12}-\left(\frac{132}{12}\right)^2 \\
& =141-121 \\
& =20
\end{aligned}$
$\therefore \quad$ Standard deviation $=\sqrt{20}=2 \sqrt{5}$

Topic: Mathematics, Statistics, MHT CET, MHT CET 2023 (09 May Shift 1)
The standard deviation of the following distribution is
$\begin{array}{|l|c|c|c|c|}
\hline \begin{array}{l}
\text { Class interval } \\
\end{array} & 0-10 & 10-20 & 20-30 & 30-40 \\
\hline \begin{array}{l}
\text { Frequency } \\
\end{array} & 1 & 3 & 4 & 2 \\
\hline
\end{array}$

Options:
A 9
B 8
C 7
D 6

The correct answer is:
9

Solution:
No solution. Refer to answer key.

Topic: Mathematics, Statistics, AP EAMCET, AP EAMCET 2017 (24 Apr Shift 2)
The standard deviation of the numbers 22, 26, 28, 20, 24, 30 is

Options:
A 2
B 2.4
C 3.24
D 3.42

The correct answer is:
3.42

Solution:
We know that

S.D=∑xi-μ2N

where,

μ→Mean

N→Number of terms

xi→terms

Now,

μ=22+26+28+20+24+306

⇒μ=1506=25

Now,

xi	xi-μ2

22	22-252=9

26	26-252=1

28	28-252=9

20	20-252=25

24	24-252=1

30	30-252=25

Therefore, ∑xi-μ2=70

S.D=706=3.42

Topic: Mathematics, Statistics, AP EAMCET, AP EAMCET 2018 (25 Apr Shift 1)
The standard deviation of the numbers 22, $26,28,20,24,30$ is

Options:
A 2
B 2.4
C 3.24
D 3.42

The correct answer is:
3.42

Solution:
Given numbers 22, 26, 28, 20, 24, 30
$$
\bar{x}=\frac{22+26+28+20+24+30}{6}=\frac{150}{6}=25
$$
Now, $\left(x_i-\bar{x}\right)^2$ is $9,1,9,25,1,25$
$$
\begin{aligned}
& \text { Now, } S D=\sqrt{\frac{\Sigma\left(x_i-\bar{x}\right)^2}{n}} \\
& =\sqrt{\frac{9+1+9+25+1+25}{6}}=\sqrt{\frac{70}{6}}=3.42
\end{aligned}
$$

Topic: Mathematics, Statistics, AP EAMCET, AP EAMCET 2018 (23 Apr Shift 1)
The standard deviation of the numbers 31 , $32,33 \ldots 46,47$ is

Options:
A $\sqrt{\frac{17}{12}}$
B $\sqrt{\frac{47^{2}-1}{12}}$
C $2 \sqrt{6}$
D $4 \sqrt{3}$

The correct answer is:
$2 \sqrt{6}$

Solution:
Given, numbers : $31,32,33, \ldots, 46,47$. The standard deviation of $31,32,33, \ldots, 47$ will be the same as those of $1,2,3, \ldots, 17$. (decreasing each item by 30 )
$\because \quad \mathrm{SD}=\sqrt{\frac{n^{2}-1}{12}}$
Here, $\quad n=17$
$\begin{aligned}
\therefore \quad \mathrm{SD} &=\sqrt{\frac{17^{2}-1}{12}}=\sqrt{\frac{288}{12}} \\
&=\sqrt{24}=2 \sqrt{6}
\end{aligned}$

Topic: Mathematics, Statistics, KCET, KCET 2021
The standard deviation of the observations $5,5,5,5,5$ is

Options:
A 0
B 5
C 20
D 25

The correct answer is:
0

Solution:
Given observation are $5,5,5,5,5$
$\bar{x}=\frac{5+5+5+5+5}{5}=\frac{25}{5}=5$
Now, $\sigma^{2}=\sum_{i=1}^{5} \frac{\left(x_{i}-\bar{x}\right)^{2}}{N}$
Where $\mathrm{N}=$ total number of observations.
Variance $=\sigma^{2}$
$\frac{(5-5)+(5-5)+(5-5)+(5-5)+(5-5)}{=}$
Hence, standard deviation $=\sqrt{v a r}=0$

Topic: Mathematics, Statistics, NDA, NDA 2013 (Phase 1)
The standard deviation of the scores 505, 510, $515,520, \ldots \ldots ., 595$ is

Options:
A $500+5 \sqrt{30}$
B $505+\sqrt{30}$
C $5 \sqrt{30}$
D $5+\sqrt{30}$

The correct answer is:
$5 \sqrt{30}$

Solution:
Given, scores 505, 510, 515, 520, ... 595 the mean of these 19 scores $=550=\bar{X}$
So,
$$
\begin{aligned}
\sum_{i=1}^{19}\left(x_i-\bar{x}\right)^2 & =2\left(45^2+40^2+35^2+\ldots+5^2\right) \\
& =2 \times 5^2 \times\left[1^2+2^2+3^2+\ldots+9^2\right]
\end{aligned}
$$
So, standard derivation
$$
=\sqrt{\frac{\sum_{i=1}^{19}\left(x_i-\bar{x}\right)^2}{n}}=\sqrt{\frac{2 \times 5^2 \times \frac{9 \times 10 \times 19}{6}}{19}}=5 \sqrt{30}
$$

Topic: Mathematics, Statistics, AP EAMCET, AP EAMCET 2018 (23 Apr Shift 2)
The standard deviations of two sets of observations $X=\left\{x_i\right\}$ and $Y=\left\{y_i\right\}$ $(i=1,2, \ldots, 100)$ are respectively 5 and 6 . If $\bar{x}, \bar{y}$ are their means and $\sum_{i=1}^{100}\left(x_i-\bar{x}\right)\left(y_i-\bar{y}\right)=600$, then the standard deviation of $Z=\left\{z_i / z_i=x_i-y_i\right)$ is

Options:
A 12
B 6
C 7
D 10

The correct answer is:
7

Solution:
From the given information, we can say that
$\sqrt{\frac{\sum_{i=1}^{100}\left(x_i-\bar{x}\right)^2}{100}}=5 \text { and } \sqrt{\frac{\sum_{i=1}^{100}\left(y_i-\bar{y}\right)^2}{100}}=6$
$\therefore \quad \sum_{i=1}^{100}\left(x_i-\bar{x}\right)^2=2500$ and $\sum_{i=1}^{100}\left(y_i-\bar{y}\right)^2=3600$
and it is also given that
$\sum_{i=1}^{100}\left(x_i-\bar{x}\right)\left(y_i-\bar{y}\right)=600$
$\because \quad a^2+b^2-2 a b=(a-b)^2$
$\sum_{i=1}^{100}\left(x_i-\bar{x}\right)^2+\sum_{i=1}^{100}\left(y_i-\bar{y}\right)^2-\sum_{i=1}^{100}\left(x_i-\bar{x}\right)\left(y_i-\bar{y}\right)$
$=\sum_{i=1}^{100}\left(\left(x_i-\bar{x}\right)-\left(y_i-\bar{y}\right)\right)^2$
$\Rightarrow \quad \sum_{i=1}^{100}\left[\left(x_i-\bar{x}\right)-\left(y_i-\bar{y}\right)\right]^2=2500+3600-1200$
$\Rightarrow \quad \sum_{i=1}^{100}\left[\left(x_i-y_i\right)-(\bar{x}-\bar{y})\right]^2=4900$
$\Rightarrow \sqrt{\frac{\sum_{i=1}^{100}\left[\left(x_i-y_i\right)-(\bar{x}-\bar{y})\right]^2}{100}}=7$

Topic: Mathematics, Statistics, TS EAMCET, TS EAMCET 2020 (11 Sep Shift 2)
The standard deviations of xii=1,2,….10 and yii=1,…10 are respectively a and b. x¯, y¯ are the means of these two sets of observations respectively. If zi=xi-x¯yi-y¯ and ∑i=110zi=c, then the standard deviation of the observations xi-yi, i=1,2,…10 is

Options:
A a2+b2+c5
B a2+b2-c5
C a2+b2-c25
D a2+b2+c25

The correct answer is:
a2+b2-c5

Solution:
It is given that, zi=xi-x¯yi-y¯

And ∑i=110zi=c

Rewrite the above equation.

∑i=110xi-x¯yi-y¯=c

⇒ ∑i=110xiyi-xiy¯-x¯yi+x¯·y¯=c

⇒ ∑i=110xiyi-∑i=110xiy¯-∑i=110x¯yi+∑i=110x¯·y¯=c

⇒ ∑i=110xiyi-10x¯·y¯-10x¯·y¯+10x¯·y¯=c

⇒ ∑i=110xiyi-10x¯·y¯=c

Now,

σx1-y1=110∑xi-yi2-x¯-y¯2

=110∑xi2+110∑yi2-15∑xiyi-x¯2-y¯2+2xy¯

=110∑xi2-x¯2+110∑yi2-y¯2-15∑xiyi-10xy¯

=a2+b2-c5

Topic: Mathematics, Statistics, AP EAMCET, AP EAMCET 2019 (21 Apr Shift 2)
The standard $E_{\text {red }}^{\circ}$ values of $A, B$ and $C$ are $+0.68 \mathrm{~V},-2.54 \mathrm{~V},-0.50 \mathrm{~V}$ respectively. The
order of their reducing power is

Options:
A $A>B>C$
B $A>C>B$
C $C>B>A$
D $B>C>A$

The correct answer is:
$B>C>A$

Solution:
Reducing character is based upon higher negative value of reduction electrode potential.
Thus, order of reducing character is
$$
B>C>A
$$

Topic: Chemistry, Electrochemistry, MHT CET, MHT CET 2007
The standard electrode potential $\left(\mathrm{E}^{-}\right)$values of $\left(\mathrm{Al}^{3+} / \mathrm{Al}, \mathrm{Ag}^{+} / \mathrm{Ag}, \mathrm{K}^{+} / \mathrm{K}\right.$ and $\mathrm{Cr}^{3+} / \mathrm{Cr}$ are $-1.66 \mathrm{~V}, 0.80 \mathrm{~V}, 2.93 \mathrm{~V}$ and $-0.74 \mathrm{~V}$, respectively. The correct decreasing order of reducing power of the metal is

Options:
A $\mathrm{Ag}>\mathrm{Cr}>\mathrm{Al}>\mathrm{K}$
B $\mathrm{K}>\mathrm{Al}>\mathrm{Cr}>\mathrm{Ag}$
C $\mathrm{K}>\mathrm{Al}>\mathrm{Ag}>\mathrm{Cr}$
D $\mathrm{Al}>\mathrm{K}>\mathrm{Ag}>\mathrm{Cr}$

The correct answer is:
$\mathrm{K}>\mathrm{Al}>\mathrm{Cr}>\mathrm{Ag}$

Solution:
More negative the value of standard reduction potential, higher is the reduction power.
i.e. Reducing power $\propto \frac{1}{\text { standard reduction potential }}$
Thus, the correct decreasing order of reducing power of the metal is
$$
\begin{aligned}
& \begin{array}{llll}
\mathrm{K} & > & \mathrm{Al} & >
\end{array} \\
& \left(\mathrm{E}_{\mathrm{K}^{+} / \mathrm{K}}^0=-2.93 \mathrm{~V}\right) \quad\left(\mathrm{E}_{\mathrm{Al}^{3+} / \mathrm{Al}}^0=-1.66 \mathrm{~V}\right) \\
& \mathrm{Cr}>\mathrm{Ag} \\
& \left(\mathrm{E}_{\mathrm{Cr}^{3+} / \mathrm{Cr}}^0=-0.74 \mathrm{~V}\right) \quad\left(\mathrm{E}_{\mathrm{Ag}^{+}}^0 / \mathrm{Ag}=0.80 \mathrm{~V}\right) \\
&
\end{aligned}
$$

Topic: Chemistry, Electrochemistry, NEET, NEET 2019 (Odisha)
The standard electrode potential for Daniell cell is $1.1 \mathrm{~V}$. What is the standard Gibbs energy for the reaction?

Options:
A $212.3 \mathrm{~kJ} \mathrm{~mol}^{-1}$
B $-2123 \mathrm{~kJ} \mathrm{~mol}^{-1}$
C $106.15 \mathrm{~kJ} \mathrm{~mol}^{-1}$
D $-106.15 \mathrm{~kJ} \mathrm{~mol}^{-1}$

The correct answer is:
$-2123 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution:
Net reaction occuring in a Daniell cell is
$$
\begin{aligned}
\mathrm{Zn}(s) &+\mathrm{Cu}^{2+}(a q) \longrightarrow \mathrm{Zn}^{2+}(a q)+\mathrm{Cu}(s) \\
\Delta G &=-n F E^{\circ}=-2 \times 96500 \times 1.1 \\
&=-212300 \mathrm{~J} \mathrm{~mol}{ }^{-1}=-212.3 \mathrm{~kJ} \mathrm{~mol}^{-1}
\end{aligned}
$$

Topic: Chemistry, Electrochemistry, COMEDK, COMEDK 2015
The standard electrode potential for the half-cell reactions are
$$
\begin{aligned}
&\mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn} ; \mathrm{E}^{\circ}=-0.76 \mathrm{~V} \\
&\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe} ; \mathrm{E}^{\circ}=-0.44 \mathrm{~V}
\end{aligned}
$$
The emf of the cell reaction,
$\mathrm{Fe}^{2+}+\mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+}+\mathrm{Fe}$ is

Options:
A $-0.32 \mathrm{~V}$
B $-1.20 \mathrm{~V}$
C $+1.20 \mathrm{~V}$
D $+0.32 \mathrm{~V}$

The correct answer is:
$+0.32 \mathrm{~V}$

Solution:
$\mathrm{Zn}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Zn} ; \mathrm{E}^{\circ}=-0.76 \mathrm{~V}$ $\mathrm{Fe}^{2+}+2 \mathrm{e}^{-} \longrightarrow \mathrm{Fe} ; \mathrm{E}^{\circ}=-0.44 \mathrm{~V}$
Cell reaction is
$\begin{aligned} \mathrm{Fe}^{2+}+\mathrm{Zn} \longrightarrow \mathrm{Zn}^{2+}+\mathrm{Fe} \\ \mathrm{E}_{\text {cell }} &=\mathrm{E}_{\text {cathode }}-\mathrm{E}_{\text {anode }} \\ &=-0.44-(-0.76) \\ &=-0.44+0.76 \\ &=0.32 \mathrm{~V} \end{aligned}$

Topic: Chemistry, Electrochemistry, KCET, KCET 2009
The standard electrode potential (in V) values for $\mathrm{Al}^{3+}$ / $\mathrm{Al}, \mathrm{Tl}^{3+} / \mathrm{T} 1$ are respectively

Options:
A $-1.66,-1.26$
B $+1.66,+1.26$
C $-1.66,+1.26$
D $+1.66,-1.26$

The correct answer is:
$-1.66,+1.26$

Solution:
($\mathrm{Al}^{3+}$ is much more stable than $\mathrm{Al}$ and due to electropositive nature, the standard reduction potential of $\mathrm{Al}^{3+} / \mathrm{Al}$ is negative.
$\mathrm{Tl}$ is not as electropositive as $\mathrm{Al}$ and therefore $\mathrm{Tl}^{3+}$ can easily be reduced to Tl because of which its SEP value is positive.

Topic: Chemistry, Electrochemistry, AP EAMCET, AP EAMCET 2023 (18 May Shift 2)
The standard electrode potential of $\mathrm{Zn}^{2+} / \mathrm{Zn}$ is $-0.76 \mathrm{~V}$ and that of $\mathrm{Ca}^{2+} / \mathrm{Cu}$ is $0.34 \mathrm{~V}$. The emf $(\mathrm{V})$ and the free energy change $\left(\mathrm{kJ} \mathrm{mol}^{-1}\right)$, respectively, for a Daniel cell will be

Options:
A $-0.42$ and 81
B $1.1$ and -213
C $-1.1$ and 213
D $0.42$ and $-81$

The correct answer is:
$1.1$ and -213

Solution:
$\mathrm{E}_{\mathrm{Cell}}^{0}=\mathrm{E}^{0} \mathrm{Cu}^{2+}\left|\mathrm{Cu}+\mathrm{E}^{0} \mathrm{Zn}\right| \mathrm{Zn}^{2+}$
$\mathrm{E}_{\mathrm{cell}}^{0}=0.34+0.76=+1.1 \mathrm{~V}$
$\Delta G=-2 \times 96500 \times 1.1=-213 V$

Topic: Chemistry, Electrochemistry, KVPY, KVPY 2013 (SB/SX)
The standard electrode potentials of

Zn2+ | Zn , Cu2+ | Cu and Ag+ | Ag are respectively - 0.76, 0.34 and 0.8 V. The following cells were constructed

I - Zn | Zn2+ || Cu2+ | Cu

II - Zn | Zn2+ || Ag+ | Ag

III - Cu | Cu2+ || Ag+ | Ag

What is the correct order of EcellÂ° of these cells?

Options:
A II > III > I
B II > I > III
C I > II > III
D III > I > II

The correct answer is:
II > I > III

Solution:
Given,
Zn 2+ â+âââ 2e â â Zn , E Â° =â0.76 V
Cu 2+ +ââ 2e â â Cu , E Â° = 0.34 V
Ag + +ââ e â â Ag , E Â° = 0.8 V
Cell reaction of (I) is
Zn + Cu2+ â Zn2+ + Cu
Ecello=Eredo(cathode)-Eredo(anode)
=0.34â â0.76 =1.10âV
Cell reaction of (II) is
Zn + 2Ag + â Zn 2+ + 2Ag
E Cell o =+0.80â â0.76 =+1.56âV
Cell reaction of (III) is
Cu + 2Ag + â Cu 2+ + 2Ag
E Cell o =+0.80â 0.34 =+0.46âV
So, the correct order of EcellÂ° of these cell is
II > I > III.

Topic: Chemistry, Electrochemistry, NEET
The standard e.m.f of a cell, involving one electron change is found to be 0.591 V at 25 oC . The equilibrium constant of the reaction is F=96500 Cmol-1,R=8.314 JK-1mol-1

Options:
A 1.0×1030
B 1.0×1015
C 1.0×105
D 1.0×1010

The correct answer is:
1.0×1010

Solution:
Relation between Keq and Ecell is
E cell o = 2.303RT nF logK eq
E cell o = 0.0591 n logK eq
0.591= 0.0591 1 logK eq
logK eq =10
K eq =1× 10 10

Topic: Chemistry, Electrochemistry, NEET
The standard emf of a galvanic cell involving 2 moles of electrons in its redox reaction is $0.59 \mathrm{~V}$. The equilibrium constant for the redox reaction of the cell is

Options:
A $10^{20}$
B $10^{5}$
C 10
D $10^{10}$

The correct answer is:
$10^{20}$

Solution:
In a galvanic cell, $\Delta G^{\circ}=-n F E^{\circ}$
Relationship between $\Delta G^{\circ}$ and equilibrium constant
$$
\Delta G^{\circ}=-2.303 R T \log K_{p}
$$
From Eqs. (i) and (ii), we get
$$
\begin{gathered}
-2.303 R T \log K_{p}=-n F E^{\circ} \\
\log K_{p}=\frac{n F E^{\circ}}{2.303 R T}=\frac{n E^{\circ}}{\frac{2.303 R T}{F}} \\
=\frac{2 \times 0.59}{0.059}=20 \\
K_{p}=\mathrm{AL}(20) \quad(\mathrm{AL}=\text { Antilog }) \\
K_{p}=10^{20}
\end{gathered}
$$

Topic: Chemistry, Electrochemistry, KCET, KCET 2011
The standard e.m.f of a galvanic cell involving cell reaction with $\mathrm{n}=2$ is found to be $0.295 \mathrm{~V}$ at $25^{\circ} \mathrm{C}$. The equilibrium constant of the reaction would be:
(Given $\mathrm{F}=96500 \mathrm{C} \mathrm{mol}^{-1}, \mathrm{R}=8.314 \mathrm{~J}$ $\mathrm{K}^{-1} \mathrm{~mol}^{-1}$

Options:
A $2.0 \times 10^{11}$
B $4.0 \times 10^{12}$
C $1.0 \times 10^2$
D $1.0 \times 10^{10}$

The correct answer is:
$1.0 \times 10^{10}$

Solution:
$\mathrm{E}=\mathrm{E}^{\circ}-\frac{0.0591}{n} \log _{10} \mathrm{Q}$ at $25^{\circ} \mathrm{C}$
At equilibrium, $\mathrm{E}=0, \mathrm{Q}=\mathrm{K}$
$$
0=\mathrm{E}^{\circ}-\frac{0.0591}{n} \log _{10} \mathrm{~K}
$$
or, $\mathrm{K}=$ Antilog $\left[\frac{n \mathrm{E}^{\circ}}{0.0591}\right]$
or, $\mathrm{K}=$ Antilog $\left[\frac{2 \times 0.295}{0.0591}\right]$
$$
\begin{aligned}
& =\text { Antilog }\left[\frac{0.590}{0.0591}\right] \\
& =\text { Antilog } 10=1 \times 10^{10} .
\end{aligned}
$$
Related Theory
The cell potential is the difference between the electrode potentials (reduction potentials) of the cathode and anode. It is called the standard electromotive force (emf) of the cell when no current is drawn through the cells.

Topic: Chemistry, Electrochemistry, NEET, NEET 2004
The standard e.m.f. of a galvanic cell involving cell reaction with \(n=2\) is found to be \(0.295 \mathrm{~V}\) at \(25^{\circ} \mathrm{C}\). The equilibrium constant of the reaction would be (Given \(\mathrm{F}=96500 \mathrm{C} \mathrm{mol}^{-1}\); \(\mathrm{R}=8.314 \mathrm{~J}^{-1} \mathrm{~mol}^{-1}\))

Options:
A \(2.0 \times 10^{11}\)
B \(4.0 \times 10^{12}\)
C \(1.0 \times 10^2\)
D \(1.0 \times 10^{10}\)

The correct answer is:
\(1.0 \times 10^{10}\)

Solution:
\(\mathrm{E}^{\circ}=\frac{0.0591}{\mathrm{n}} \log _{10} \mathrm{~K}\)
Here, \(\mathrm{n}=2, \mathrm{E}^{\circ}=0.295\)
\(\therefore \log _{10} \mathrm{~K}=\frac{2 \times 0.295}{0.0591}=10 \text { or } \mathrm{K}=10^{10}\)

Topic: Chemistry, Electrochemistry, VITEEE, VITEEE 2022
The standard emf of galvanic cell involving 3 moles of electrons in its redox reaction is $0.59 \mathrm{~V}$. The equilibrium constant for the reaction of the cell is

Options:
A $10^{25}$
B $10^{20}$
C $10^{15}$
D $10^{30}$

The correct answer is:
$10^{30}$

Solution:
$\begin{aligned} \mathrm{E}_{\text {cell }}^{\circ} &=\frac{0.0591}{\mathrm{n}} \log \mathrm{K}_{\mathrm{c}} \\ 0.59 &=\frac{0.0591}{3} \log \mathrm{K}_{\mathrm{c}} \\ \frac{0.59 \times 3}{0.059} &=\log \mathrm{K}_{\mathrm{c}} \\ \therefore \quad \log \mathrm{K}_{\mathrm{c}} &=30 \\ \mathrm{~K}_{\mathrm{c}} &=\text { antilog } 30 \\ &=10^{30} \end{aligned}$

Topic: Chemistry, Chemical Equilibrium, KCET, KCET 2010
The standard emf of the cell $\mathrm{Zn}\left|\mathrm{Zn}^{2+}\right|\left|\mathrm{Ag}^{+}\right| \mathrm{Ag}$ is $1.56 \mathrm{~V}$. If the standard reduction potential of $\mathrm{Ag}$ is $0.8 \mathrm{~V}$, the standard oxidation potential of $\mathrm{Zn}$ is

Options:
A $-0.76 \mathrm{~V}$
B $+0.76 \mathrm{~V}$
C $-2.36 \mathrm{~V}$
D $+2.36 \mathrm{~V}$

The correct answer is:
$+0.76 \mathrm{~V}$

Solution:
For the cell, $\mathrm{Zn}\left|\mathrm{Zn}^{2+} \| \mathrm{Ag}^{+}\right| \mathrm{Ag}$
$E_{\text {cell }}^{\circ}=E_{\text {cathode }}^{\circ}-E_{\text {anode }}^{\circ}$
$1.56 \mathrm{~V}=E_{\mathrm{Ag}^{+} / \mathrm{Ag}}^{0}-E_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{0}$
$E_{\mathrm{Zn}^{2+} / \mathrm{Zn}}^{\circ}=-0.76 \mathrm{~V}$
$\therefore \quad E_{\mathrm{Zn} / \mathrm{Zn}^{2+}}^{\circ}=-E_{\mathrm{Zn}^{2}+/ \mathrm{Zn}}^{\circ}$
$=-(-0.76)=+0.76 \mathrm{~V}$

Topic: Chemistry, Electrochemistry, COMEDK, COMEDK 2014
The standard enthalpies of formation of $A\left(\mathrm{NH}_3\right), B\left(\mathrm{CO}_2\right), C(\mathrm{HI})$ and $D\left(\mathrm{SO}_2\right)$ are -46.19 , $-393.4,+24.94$ and $-296.9 \mathrm{~kJ} \mathrm{~mol}^{-1}$ respectively. The increasing order of their stability is

Options:
A $B \lt D \lt A \lt C$
B $C \lt A \lt D \lt B$
C $D \lt B \lt C \lt A$
D $A \lt C \lt D \lt B$

The correct answer is:
$C \lt A \lt D \lt B$

Solution:
Lower the value of standard enthalpy of formation, higher the stability. The increasing order of standard enthalpies of formation of given molecules is
Thus, the increasing order of stability is
$C \lt A \lt D \lt B$

Topic: Chemistry, Thermodynamics (C), JIPMER, JIPMER 2014
The standard enthalpies of formation of CO2(g),H2O(l) and glucose (s) at 25Â°C are â400 kJ/mol, â300 kJ/mol and â1300 kJ/mol , respectively. The standard enthalpy of combustion per gram of glucose at 25Â°C is

Options:
A + 2900 kJ
B −2900 kJ
C −16.11 kJ
D + 16.11 kJ

The correct answer is:
−16.11 kJ

Solution:
C 6 H 12 O 6 +6 CO 2 →6 O 2 +6 H 2 O
CO2→-400 kJ/mol
H2O=-300 kJ/mol
Glucose =-1300 kJ/mol
ΔH=ΔHP-ΔHR=6ΔHf(CO2)+6Hf(H2O)-ΔHf(C6H12O6)
=6×-400+6-300--1300=-2400-1800+1300=-2900 kJ/mol
ΔH=-2900180=-16.11 kJ/g

Topic: Chemistry, Thermodynamics (C), NEET