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

The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. $A$ simple model for such a concentration cell involving a metal $M$ is:
$M_{(s)} \mid M^{+}(aq; 0.05 \ M) \parallel M^{+}(aq; 1 \ M) \mid M_{(s)}$
For the above electrolytic cell the magnitude of the cell potential $|E_{cell}|=70 \ mV$.
$1.$ For the above cell
$(A)$ $E_{cell} < 0 ; \Delta G > 0$ $(B)$ $E_{cell} > 0 ; \Delta G < 0$
$(C)$ $E_{cell} < 0 ; \Delta G^{\circ} > 0$ $(D)$ $E_{cell} > 0 ; \Delta G^{\circ} > 0$
$2.$ If the $0.05 \ M$ solution of $M^{+}$ is replaced by $0.0025 \ M$ $M^{+}$ solution,then the magnitude of the cell potential would be
$(A)$ $35 \ mV$ $(B)$ $70 \ mV$ $(C)$ $140 \ mV$ $(D)$ $700 \ mV$
Give the answer for questions $1$ and $2$.

Give the $E_{cell}$ equation for the general reaction: $aA + bB \to cC + dD$.

$H_{2(g)} + 2 AgCl_{(s)} \rightleftharpoons 2 Ag_{(s)} + 2 HCl_{(aq)}$. The $E^{\circ}_{cell}$ at $25^{\circ} C$ for the cell is $0.22 \ V$. The equilibrium constant at $25^{\circ} C$ is

Calculate the equilibrium constant of the following reaction:
$\text{Cu(s)} + 2\text{Ag}^+_{\text{(aq)}} \rightarrow \text{Cu}^{2+}_{\text{(aq)}} + 2\text{Ag(s)}$,given $E^\circ_{\text{cell}} = 0.46 \text{ V}$.

At $298 \, K$,the standard reduction potential for $Cu^{2+}/Cu$ electrode is $0.34 \, V$.
Given: $K_{sp} \text{ of } Cu(OH)_2 = 1 \times 10^{-20}$
Take $\frac{2.303 RT}{F} = 0.059 \, V$
The reduction potential at $pH = 14$ for the above couple is $(-)x \times 10^{-2} \, V$. The value of $x$ is $........$.