Reduction of species is dependent on its reduction potential value.

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(N/A) Reduction reactions occur on the surface of the cathode. If more than one species is present near the cathode,the species with the higher $E^{\ominus}$ value undergoes reduction.
For example,in an aqueous $NaCl$ solution,both $Na^{+}$ ions from $NaCl$ and $H^{+}$ ions from $H_{2}O$ are present near the cathode.
$NaCl_{(aq)} \rightarrow Na^{+}_{(aq)} + Cl^{-}_{(aq)}$
$H_{2}O_{(l)} \rightleftharpoons H^{+}_{(aq)} + OH^{-}_{(aq)}$
$(i) \ Na^{+}_{(aq)} + e^{-} \rightarrow Na_{(s)} \quad E^{\ominus} = -2.71 \ V$
$(ii) \ H^{+}_{(aq)} + e^{-} \rightarrow \frac{1}{2} H_{2(g)} \quad E^{\ominus} = 0.00 \ V$
Comparing the $E^{\ominus}$ values,reaction $(ii)$ has a higher value than reaction $(i)$. Therefore,$H^{+}$ ions from water are reduced at the cathode to produce $H_{2}$ gas.
The overall cathodic reaction is:
$H_{2}O_{(l)} + e^{-} \rightarrow \frac{1}{2} H_{2(g)} + OH^{-}_{(aq)}$
As $Na^{+}$ ions do not participate in the reaction,they remain in the solution as spectator ions,combining with $OH^{-}$ to form $NaOH$. The presence of $NaOH$ is confirmed by the pink color observed when phenolphthalein is added near the cathode.

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Using the data given below,find out the strongest reducing agent:
$E^0_{Cr_2O_7^{2-}/Cr^{3+}} = 1.33 \text{ V}$,$E^0_{Cl_2/Cl^{-}} = 1.36 \text{ V}$,$E^0_{MnO_4^-/Mn^{2+}} = 1.51 \text{ V}$,$E^0_{Cr^{3+}/Cr} = -0.74 \text{ V}$

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

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Match the following:
List-$I$List-$II$
$(A)$ Potential of hydrogen electrode at $pH = 10$$(I)$ $0.76 \ V$
$(B)$ $Cu^{2+} | Cu$$(II)$ $0.059$
$(C)$ $Zn | Zn^{2+}$$(III)$ $-0.591 \ V$
$(D)$ $\frac{2.303 RT}{F}$$(IV)$ $0.337 \ V$
$(V)$ $-0.76 \ V$

$A$ $B$ $C$ $D$
$(a)$ $(III)$ $(I)$ $(II)$ $(V)$
$(b)$ $(II)$ $(V)$ $(I)$ $(IV)$
$(c)$ $(III)$ $(IV)$ $(I)$ $(II)$
$(d)$ $(V)$ $(I)$ $(IV)$ $(II)$

The standard electrode potentials $E^{\circ} (V)$ for $Li^{+} / Li$ and $Na^{+} / Na$ respectively are:

Reduction potential of ions are given below:
$ClO_4^{-}$ $E^{\circ} = 1.19 \ V$
$IO_4^{-}$ $E^{\circ} = 1.65 \ V$
$BrO_4^{-}$ $E^{\circ} = 1.74 \ V$

The correct order of their oxidising power is:

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