For a cell reaction involving a two-electron change,the standard emf of the cell is found to be $0.295 \ V$ at $25 \ ^oC$. The equilibrium constant of the reaction at $25 \ ^oC$ will be

For the cell reaction with the indicated concentrations
$Al_{(s)} + 3Ag^{+}_{(aq)} \, (0.10 \, M) \to Al^{3+}_{(aq)} \, (0.30 \, M) + 3Ag_{(s)}$,
the measured voltage $(E_{cell})$ is $1.50 \, V$. Calculate $E^o_{cell} \, ........... \, V$
Given: $\frac{2.303 \, RT}{F} = 0.06 \, V ; \, \log \, 3 = 0.48$

What is the standard cell potential for the reaction with $K=1$ (equilibrium constant)?

For the concentration cell: $Cu | Cu^{2+} (0.01 \ M) || Cu^{2+} (0.1 \ M) | Cu$,calculate the $E_{cell}$.

Given the equilibrium constant $K_c$ of the reaction $Cu_{(s)} + 2Ag^{+}_{(aq)} \to Cu^{2+}_{(aq)} + 2Ag_{(s)}$ is $10 \times 10^{15}$,calculate the $E_{cell}^o$ of the reaction at $298 \ K$. [Given: $2.303 \ \frac{RT}{F} \text{ at } 298 \ K = 0.059 \ V$]

Calculate the $e.m.f.$ of the half-cell given below:
$Fe | FeSO_4$ $(a = 0.1 \ M)$
Given: $E^o_{OP} = 0.44 \ V$ (in $V$)