Calculate the cell potential for the cell $Zn_{(s)} | Zn^{2+} (0.6 \ M) || Cd^{2+} (0.85 \ M) | Cd_{(s)}$ at $298 \ K$. (Given: $E^{\circ}_{Zn^{2+}/Zn} = -0.76 \ V$ and $E^{\circ}_{Cd^{2+}/Cd} = -0.40 \ V$) (in $V$)

Calculate the emf of the half-cell given below: $Pt(s) | H_2(g, 2 \text{ atm}) | HCl(aq, 0.02 \text{ M})$, $E^\circ_{H^+/H_2} = 0 \text{ V}$. (Given: $\frac{2.303RT}{F} = 0.059$, $\log 2 = 0.3010$)

For the following electrochemical cell at $298 \ K$,
$Pt_{(s)} \mid H_2(g, 1 \ bar) \mid H^{+}(aq, 1 \ M) \parallel M^{4+}_{(aq)}, M^{2+}_{(aq)} \mid Pt_{(s)}$
$E_{\text{cell}} = 0.092 \ V$ when $\frac{[M^{2+}_{(aq)}]}{[M^{4+}_{(aq)}]} = 10^x$
Given : $E^0_{M^{4+}/M^{2+}} = 0.151 \ V$; $2.303 \frac{RT}{F} = 0.059 \ V$
The value of $x$ is

What is the change in the reduction potential of a hydrogen electrode when the $pH$ of the initial solution changes from $0$ to $7$ by neutralization?

The equilibrium constant of the reaction $Cu_{(s)} + 2Ag^{+}_{(aq)} \to Cu^{2+}_{(aq)} + 2Ag_{(s)}$ with $E^{\circ} = 0.46 \ V$ at $298 \ K$ is approximately:

The equilibrium constant for the following general reaction is $10^{30}$. Calculate $E^{o}$ for the cell at $298 \ K$ ............ $V$
$2X_{2(s)} + 3Y^{2+}_{(aq)} \to 2{X_{2}}^{3+}_{(aq)} + 3Y_{(s)}$