Calculate the equilibrium constant $(K_C)$ for the cell obtained by connecting two electrodes with standard electrode potentials $E^o_{(Sn^{2+}|Sn)} = -0.14 \ V$ and $E^o_{(Ni^{2+}|Ni)} = -0.23 \ V$ at $298 \ K$.

Calculate $E_{cell}$ of the reaction in $V$:
$Mg_{(s)} + 2Ag^{+}(0.0001 \ M) \to Mg^{+2}(0.100 \ M) + 2Ag_{(s)}$
Given that $E_{cell}^o = 3.17 \ V$.

An electrochemical cell consists of the following two redox couples, $M^{x+}(aq)/M(s)$ $[E^{\ominus}_{red} = +0.15 \text{ V}]$ and $Fe^{3+}(aq)/Fe(s)$ $[E^{\ominus}_{red} = -0.036 \text{ V}]$. The cell $EMF$ is recorded to be $0.2057 \text{ V}$. If the reaction quotient of the electrochemical reaction is found to be $10^{-2}$, then the value of $x$ is . . . . . . . (Nearest integer) [Given: $M$ is a $p$-block metal and $\frac{2.303RT}{F} = 0.059 \text{ V}$]

For a reaction,$A + B^{2+} \to B + A^{2+}; E^{\circ} = 0.2955 \ V$. Hence,the equilibrium constant of the reaction at $25 \ ^oC$ is:

For a cell reaction involving two electron changes,$E_{\text{cell}}^{\circ} = 0.3 \text{ V}$ at $25^{\circ}\text{C}$. The equilibrium constant of the reaction is:

Write the Nernst equation and calculate the $emf$ of the following cells at $298 \, K$:
$(i) \; Mg_{(s)} | Mg^{2+}(0.001 \, M) || Cu^{2+}(0.0001 \, M) | Cu_{(s)}$
$(ii) \; Fe_{(s)} | Fe^{2+}(0.001 \, M) || H^{+}(1 \, M) | H_{2(g)}(1 \, bar) | Pt_{(s)}$
$(iii) \; Sn_{(s)} | Sn^{2+}(0.050 \, M) || H^{+}(0.020 \, M) | H_{2(g)}(1 \, bar) | Pt_{(s)}$
$(iv) \; Pt_{(s)} | Br_{2(l)} | Br^{-}(0.010 \, M), H^{+}(0.030 \, M) || H_{2(g)}(1 \, bar) | Pt_{(s)}$