Nernst Equation

(A) \underset{\text{ 0.05/2 mole}}{ { PbSO4(s) }} \underset{\text{ 0.05F}}{ { 2e^{-}}} (C) \underset{\text{ 0.05/2 mole 0.05 F}}{ { PbSO4 + 2e^{-}}}

The correct answer is 275.
$X+Y^{2+}\to Y+X^{2+}$
$E_{\text{Cell}}^0=0.36-(-2.36)=2.72V$
$E_{\text{Cell}}=2.72-\frac{0.06}{2}\log \frac{0.001}{0.01}$
$=2.72+0.03=2.75V$
$=275\times 10^{-2}V$

The standard Gibbs free energy change is given by:

The equation for the cell potential is:

For the reaction:

The net reaction is:

Using the Nernst equation:

Thus, by using the equation and solving for the logarithm of the equilibrium constant , we get:

This result shows the relationship between the standard cell potential and the equilibrium constant. A higher value indicates a more spontaneous reaction, which can be derived from the Nernst equation.

The correct answer is 3.
Overall reaction :-
$H_{2(g)}+M_{\text{(aq)}}^{3+}>M_{\text{(aq)}}^{+}+2H_{\text{(aq)}}^{+}$
$E_{\text{Cell}}=E_{\text{Cathode}}^o-E_{\text{anode}}^{\circ }-\frac{0.059}{2}\log \frac{\left[M^{+}\right]\times 1^2}{\left[M^{+3}\right]1}$
$0.1115=0.2-\frac{0.059}{2}\log \frac{\left[M^{+}\right]}{\left[M^{+3}\right]}$
$3=\log \frac{\left[M^{+}\right]}{\left[M^{+3}\right]}$
$\therefore a=3$