Given the standard electrode potentials,$K^{+}/K = -2.93 \, V$,$Ag^{+}/Ag = 0.80 \, V$,$Hg^{2+}/Hg = 0.79 \, V$,$Mg^{2+}/Mg = -2.37 \, V$,and $Cr^{3+}/Cr = -0.74 \, V$,arrange these metals in their increasing order of reducing power.

The $emf$ of a galvanic cell constituted with the electrodes $Zn^{2+} | Zn$ $(-0.76 \ V)$ and $Fe^{2+} | Fe$ $(-0.41 \ V)$ is

Calculate the standard cell potential $(E_{cell}^{o})$ for the following electrochemical cells:
$(i)$ $Al_{(s)}|Al_{(1M)}^{3+}||Cu_{(1M)}^{2+}|Cu_{(s)}$
$(ii)$ $Al_{(s)}|Al_{(1M)}^{3+}||Zn_{(1M)}^{2+}|Zn_{(s)}$
$(iii)$ $Al_{(s)}|Al_{(1M)}^{3+}||Ag_{(1M)}^{+}|Ag_{(s)}$
$($ Given: $E_{Al^{3+}|Al}^{o} = -1.66 \ V$,$E_{Zn^{2+}|Zn}^{o} = -0.76 \ V$,$E_{Cu^{2+}|Cu}^{o} = 0.34 \ V$,$E_{Ag^{+}|Ag}^{o} = 0.80 \ V$ $)$

Given:
$Co^{3+} + e^- \longrightarrow Co^{2+}; E^o = 1.81 \ V$
$Pb^{4+} + 2e^- \longrightarrow Pb^{2+}; E^o = + 1.67 \ V$
$Ce^{4+} + e^- \longrightarrow Ce^{3+}; E^o = + 1.61 \ V$
$Bi^{3+} + 3e^- \longrightarrow Bi; E^o = + 0.20 \ V$
Oxidizing power of the species will increase in the order:

The standard reduction potentials at $298 \ K$ for the following half-reactions are given:
$Zn^{2+}_{(aq)} + 2e^- \rightleftharpoons Zn_{(s)}; E^\circ = -0.762 \ V$
$Cr^{3+}_{(aq)} + 3e^- \rightleftharpoons Cr_{(s)}; E^\circ = -0.740 \ V$
$2H^+_{(aq)} + 2e^- \rightleftharpoons H_{2(g)}; E^\circ = 0.00 \ V$
$Fe^{3+}_{(aq)} + e^- \rightleftharpoons Fe^{2+}_{(aq)}; E^\circ = 0.770 \ V$
Which is the strongest reducing agent?

What is the procedure to determine the cell potential?