The standard Gibbs energy change for the Daniel cell reaction is $Zn_{(s)} + Cu^{2+}_{(aq)} \longrightarrow Zn^{2+}_{(aq)} + Cu_{(s)}$ where $E_{\text{cell}}^{\circ} = 1.1 \ V$.

Given that $E_{O_2/H_2O}^o = +1.23 \ V$; $E_{S_2O_8^{2-}/SO_4^{2-}}^o = 2.05 \ V$; $E_{Br_2/Br^-}^o = +1.09 \ V$; $E_{Au^{3+}/Au}^o = 1.4 \ V$. The strongest oxidizing agent is

If the half-cell reaction $A + e^- \to A^-$ has a large negative reduction potential,it follows that

The standard electrode potentials $(E^{0}_{Red})$ of four elements $A, B, C,$ and $D$ are $-3.05 \ V, -1.66 \ V, -0.40 \ V,$ and $0.80 \ V$ respectively. Which of the following is the most reactive?

Give the method to determine the potential of any half-cell.

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