In acidic medium,$MnO_4^-$ acts as an oxidising agent: $MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O$. If the $H^+$ ion concentration is doubled,the electrode potential of the half-cell will:

The e.m.f. of the cell in which the following reaction $Zn_{(s)} + Ni^{2+}(a = 1.0) \rightleftharpoons Zn^{2+}(a = 10) + Ni_{(s)}$ occurs,is found to be $0.5105 \ V$ at $298 \ K$. The standard e.m.f. of the cell is ............ $V$.

The $emf$ of a $Daniel$ cell at $298 \ K$ is ${E_1}$ for the cell reaction $Zn|ZnSO_4(0.01 \ M)||CuSO_4(1.0 \ M)|Cu$. When the concentration of $ZnSO_4$ is $1.0 \ M$ and that of $CuSO_4$ is $0.01 \ M$,the $emf$ changes to ${E_2}$. What is the relationship between ${E_1}$ and ${E_2}$?

$1 \ F$ electricity was passed through $Cu^{2+} (1.5 \ M, 1 \ L) / Cu$ and $0.1 \ F$ was passed through $Ag^{+} (0.2 \ M, 1 \ L) / Ag$ electrolytic cells. After this,the two cells were connected to make an electrochemical cell. The $emf$ of the cell thus formed at $298 \ K$ is:
Given: $E^0_{Cu^{2+} / Cu} = 0.34 \ V$,$E^0_{Ag^{+} / Ag} = 0.8 \ V$,$\frac{2.303 \ RT}{F} = 0.06 \ V$ (in $V$)

If the concentration of a $Zn^{2+}$ solution is diluted $10$ times,what is the change in the potential of the $Zn/Zn^{2+}$ electrode?

$2Ag^{+}_{(aq)} + Cu_{(s)} \longrightarrow Cu^{2+}_{(aq)} + 2Ag_{(s)}$
The standard potential for this reaction is $0.46 \ V$. Which change will increase the potential the most?