Question

Calculated the e.m.f. of the cell in which the following reaction takes place:
$Ni(s) + 2A{g^ + }(0.002M) \to N{i^{2 + }}(0.160M) + 2Ag(s)$
[Given that ${E_{cell}}^ \circ = 1.05V$ ]
A) $0.73$
B) $0.91$
C) $0.62$
D) $0.34$

Answer 1

In order to solve this question we first form the two equations at the anode and the cathode, after this we take the value of n by seeing the electron loss. After that with the help of Nernst equation we can calculate the induced emf.

Complete step by step solution:
In this question,
Given;
$[A{g^ + }] = 0.002M \\\ [N{i^{2 + }}] = 0.160M$
At anode there is an oxidation reaction, here Ni is oxidized with release of two electrons. And at cathode Ag is taking an electron to form Ag in solid form.
Number of loss of electrons lost is our ‘n’.
So,
$n = 2$
Here we have to find E cell, according to Nernst equation;
${E_{cell}} = {E^ \circ } - \dfrac{{0.0591}}{2}\log \dfrac{{[N{i^{2 + }}]}}{{{{[A{g^ + }]}^2}}}$
Reaction quotient is the rate of reaction of backward reaction to rate of upward reaction.
If we want it at equilibrium, we call it K.
Now putting the value of reaction quotient, we get;
${E_{cell}} = 1.05 - \dfrac{{0.0591}}{2}\log \dfrac{{0.160}}{{{{(0.002)}^2}}}$
${E_{cell}} = 1.05 - 0.02955\log (4 \times {10^4})$
${E_{cell}} = 0.91V$
This is the emf of the cell.

Note:
Nernst equation is the equation that correlates the reduction of the potential of an electrochemical reaction to the standard electrode potential.
According to Nernst equation;
${E_{cell}} = {E^ \circ } - \dfrac{{0.0591}}{2}\log \dfrac{{[N{i^{2 + }}]}}{{{{[A{g^ + }]}^2}}}$
Reaction quotient is the rate of reaction of backward reaction to rate of upward reaction.
If we want it at equilibrium, we call it K.
Here E is the reduction potential that is the measure of the tendency of the chemical that would help to acquire the electrons to the electrode so that it can reduce or oxidize easily. R is the universal gas constant, T is the temperature constant in kelvin, F is the faraday constant and Q is the reaction quotient.