Given the standard half-cell potentials $(E^{\circ})$ as: $Zn \rightarrow Zn^{2+} + 2e^{-}$; $E^{\circ} = +0.76 \ V$ and $Fe \rightarrow Fe^{2+} + 2e^{-}$; $E^{\circ} = +0.41 \ V$. Then the standard e.m.f. of the cell with the reaction $Fe^{2+} + Zn \rightarrow Zn^{2+} + Fe$ is:

$A$ $1.0 \ M$ solution with respect to each of the metal halides $AX_3, BX_2, CX_3$ and $DX_2$ is electrolysed using platinum electrodes. If
$E^o_{A^{3+}/A} = 1.50 \ V, \quad E^o_{B^{2+}/B} = 0.3 \ V,$
$E^o_{C^{3+}/C} = -0.74 \ V, \quad E^o_{D^{2+}/D} = -2.37 \ V.$
The correct sequence in which the various metals are deposited at the cathode is

Which of the following will have a standard oxidation potential less than $SHE$?

Calculate $E^o_{cell}$ for the following cell in $V$:
$Zn_{(s)} | Zn^{2+}_{(aq.)} || Ag^{+}_{(aq.)} | Ag_{(s)}$
Given: $E^o_{Zn^{2+}/Zn} = -0.76 \ V$ ; $E^o_{Ag^{+}/Ag} = 0.80 \ V$

$MnO_4^- (aq) + 8H^+ (aq) + 5e^- \to Mn^{2+} (aq) + 4H_2O (l)$; $E_1^o = 1.51 \ V$
$MnO_2 (s) + 4H^+ (aq) + 2e^- \to Mn^{2+} (aq) + 2H_2O (l)$; $E_2^o = 1.21 \ V$
$MnO_4^- (aq) + 4H^+ (aq) + 3e^- \to MnO_2 (s) + 2H_2O (l)$; $E_3^o = ?$
Value of $E_3^o$ will be ............ $V$

Assertion : $Cu^{2+}$ ions get reduced more easily than $H^{+}$ ions.
Reason : Standard electrode potential of copper is $0.34 \ V$.