In an electrochemical reaction of lead,at standard temperature,if $E^0_{(Pb^{2+}/Pb)} = m \ V$ and $E^0_{(Pb^{4+}/Pb)} = n \ V$,then the value of $E^0_{(Pb^{2+}/Pb^{4+})}$ is given by $m - x n$. The value of $x$ is $...........$. (Nearest integer)

If hydrogen electrodes dipped in two solutions of $pH=3$ and $pH=6$ are connected by a salt bridge,the $emf$ of the resulting cell is (in $V$)

$Pt_{(s)} | H_{2(g)}(1 \ bar) | H^{+}_{(aq)}(1 \ M) || M^{3+}_{(aq)}, M^{+}_{(aq)} | Pt_{(s)}$
The $E_{cell}$ for the given cell is $0.1115 \ V$ at $298 \ K$ when $\frac{[M^{+}_{(aq)}]}{[M^{3+}_{(aq)}]} = 10^{a}$.
The value of $a$ is.
Given : $E^{\circ}_{M^{3+}/M^{+}} = 0.2 \ V$
$\frac{2.303 \ RT}{F} = 0.059 \ V$

If the standard electrode potential of $Cu^{2+}/Cu$ electrode is $0.34 \, V$,what is the electrode potential of $0.01 \, M$ concentration of $Cu^{2+}$ $(T = 298 \, K)$ (in $, V$)?

The $emf$ of the following cell $Mg|Mg^{2+}(0.01 \ M)||Sn^{2+}(0.1 \ M)|Sn$ at $298 \ K$ in $V$ is: (Given: $E^{\circ}_{Mg^{2+}|Mg} = -2.34 \ V, E^{\circ}_{Sn^{2+}|Sn} = -0.14 \ V$)

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