JEE MAIN - Chemistry (2002 - No. 32)
For the following cell with hydrogen electrodes at two different pressure p1 and p2. What will be the emf for the given cell :
$$\eqalign{ & Pt({H_2})|{H^ + }(aq)|Pt({H_2}) \cr & \,\,\,\,\,{p_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,1M\,\,\,\,\,\,\,\,\,\,\,\,{p_2} \cr} $$
$$\eqalign{ & Pt({H_2})|{H^ + }(aq)|Pt({H_2}) \cr & \,\,\,\,\,{p_1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,1M\,\,\,\,\,\,\,\,\,\,\,\,{p_2} \cr} $$
$${{RT} \over F}{\log _e}{{{P_1}} \over {{P_2}}}$$
$${{RT} \over 2F}{\log _e}{{{P_1}} \over {{P_2}}}$$
$${{RT} \over F}{\log _e}{{{P_2}} \over {{P_1}}}$$
none of these
Explanation
Oxidation half cell : -
$${H_2}\left( g \right)\buildrel \, \over \longrightarrow 2{H^ + }\left( {1M} \right) + 2{e^ - }\,\,\,...{P_1}$$
Reduction half cell
$$2{H^ + }\left( {1M} \right) + 2{e^ - }\buildrel \, \over \longrightarrow {H_2}\left( g \right)\,\,\,...{P_2}$$
The net cell reaction
$$\mathop {{H_2}}\limits_{{P_1}} \left( g \right)\buildrel \, \over \longrightarrow \mathop {{H_2}}\limits_{{P_2}} \left( g \right)$$
$$E_{cell}^ \circ = 0.00\,V$$ $$\,\,\,\,\,$$ $$n=2$$
$$\therefore$$ $$\,\,\,\,$$ $${E_{cell}} = E_{cell}^ \circ - {{RT} \over {nF}}{\log _c}K$$
$$ = 0 - {{RT} \over {nF}}{\log _e}{{{P_2}} \over {{P_1}}}$$
or $$\,\,\,\,{E_{cell}} = {{RT} \over {2F}}{\log _e}{{{P_1}} \over {{P_2}}}$$
$${H_2}\left( g \right)\buildrel \, \over \longrightarrow 2{H^ + }\left( {1M} \right) + 2{e^ - }\,\,\,...{P_1}$$
Reduction half cell
$$2{H^ + }\left( {1M} \right) + 2{e^ - }\buildrel \, \over \longrightarrow {H_2}\left( g \right)\,\,\,...{P_2}$$
The net cell reaction
$$\mathop {{H_2}}\limits_{{P_1}} \left( g \right)\buildrel \, \over \longrightarrow \mathop {{H_2}}\limits_{{P_2}} \left( g \right)$$
$$E_{cell}^ \circ = 0.00\,V$$ $$\,\,\,\,\,$$ $$n=2$$
$$\therefore$$ $$\,\,\,\,$$ $${E_{cell}} = E_{cell}^ \circ - {{RT} \over {nF}}{\log _c}K$$
$$ = 0 - {{RT} \over {nF}}{\log _e}{{{P_2}} \over {{P_1}}}$$
or $$\,\,\,\,{E_{cell}} = {{RT} \over {2F}}{\log _e}{{{P_1}} \over {{P_2}}}$$
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