Consider the cell given below:
$Ag_{(s)} | Ag^{\oplus} || Cu^{2+} | Cu_{(s)}$
Given:
$Ag^{\oplus} + e^{-} \to Ag; E^{o} = x$
$Cu^{2+} + 2e^{-} \to Cu; E^{o} = y$
The value of $E^{o}_{cell}$ is:

The standard reduction potentials for $Al^{3+} | Al$,$Fe^{2+} | Fe$,and $Br_2 | Br^-$ are $-1.66 \ V$,$-0.45 \ V$,and $1.09 \ V$ respectively. What is the correct order of their reducing power?

The standard reduction potentials of $4$ elements are given below. Which of the following will be the most suitable reducing agent?
$I = -3.04 \ V$
$II = -1.90 \ V$
$III = 0 \ V$
$IV = 1.90 \ V$

Given,for $Sn^{4+} / Sn^{2+}$,standard reduction potential is $0.15 \ V$ and for $Au^{3+} / Au$,standard reduction potential is $1.5 \ V$. For the reaction,$3 Sn^{2+} + 2 Au^{3+} \longrightarrow 3 Sn^{4+} + 2 Au$,the value of $E_{\text{cell}}^{\circ}$ is:

Match the following:

List-$I$	List-$II$
$(A)$ Potential of hydrogen electrode at $pH = 10$	$(I)$ $0.76 \ V$
$(B)$ $Cu^{2+} | Cu$	$(II)$ $0.059$
$(C)$ $Zn | Zn^{2+}$	$(III)$ $-0.591 \ V$
$(D)$ $\frac{2.303 RT}{F}$	$(IV)$ $0.337 \ V$
	$(V)$ $-0.76 \ V$

$A$ $B$ $C$ $D$
$(a)$ $(III)$ $(I)$ $(II)$ $(V)$
$(b)$ $(II)$ $(V)$ $(I)$ $(IV)$
$(c)$ $(III)$ $(IV)$ $(I)$ $(II)$
$(d)$ $(V)$ $(I)$ $(IV)$ $(II)$

For the cell involving the following reaction: $Zn_{(s)} + Ni^{2+}_{(aq)} \longrightarrow Zn^{2+}_{(aq)} + Ni_{(s)}$. Given $E^{\circ}_{\text{cell}} = 0.5 \ V$. What is the standard Gibbs energy change of the cell reaction (in $kJ$)?