Consider the following electrodes $P = Zn^{2+}(0.0001 \ M) / Zn$,$Q = Zn^{2+}(0.1 \ M) / Zn$,$R = Zn^{2+}(0.01 \ M) / Zn$,$S = Zn^{2+}(0.001 \ M) / Zn$. Given $E^{\circ}(Zn^{2+} / Zn) = -0.76 \ V$,the electrode potentials of the above electrodes in volts are in the order:

Given $E^o _{Cr^{3+} / Cr} = -0.72 \, V$ and $E^o _{Fe^{2+} / Fe} = -0.42 \, V$. The potential for the cell $Cr \, | \, Cr^{3+}_{(0.1 \, M)} \, || \, Fe^{2+}_{(0.01 \, M)} \, | \, Fe$ is ......... $V$.

For the cell,$Mn_{(s)}|Mn_{(aq)}^{2+}(0.4\,M)||Sn_{(aq)}^{2+}(0.04\,M)|Sn_{(s)}$,calculate the free energy change $(\Delta G)$ at $298\,K$ in $kJ$.
Given: $E_{Mn^{2+}|Mn}^o = -1.18\,V$; $E_{Sn^{2+}|Sn}^o = -0.14\,V$; $\frac{2.303\,RT}{F} = 0.06$

Which of the following expressions indicates the correct relationship between molar conductivity of a strong electrolyte and its concentration $c$?

$Cu_{(s)} + Sn^{2+}(0.001 \ M) \rightarrow Cu^{2+}(0.01 \ M) + Sn_{(s)}$
The Gibbs free energy change for the above reaction at $298 \ K$ is $x \times 10^{-1} \ kJ \ mol^{-1}$;
The value of $x$ is ..... [nearest integer] $\left[\text{Given}: E^{\ominus}_{Cu^{2+}/Cu} = 0.34 \ V; E^{\ominus}_{Sn^{2+}/Sn} = -0.14 \ V; F = 96500 \ C \ mol^{-1}\right]$

The cell reaction involving the quinhydrone electrode is given by the following equation:
$C_6H_4(OH)_2 \rightleftharpoons C_6H_4O_2 + 2H^+ + 2e^-$,$E^{\circ} = 1.30 \ V$
What will be the electrode potential at $pH = 3$ (in $V$)?