What ratio of Pb2+ to Sn2+ concentration is needed to reverse the following cell reaction?
Sn(s)+Pb2(aq.)→Sn2+(aq.)+Pb(s)
E∘Sn2+/Sn=−0.136 volt and E∘Pb2+/Pb=−0.126 volt.
In the given cell reaction,
Ecell = E0cell − 0.0592 logSn2+Pb2+
Ecell = E0cell + 0.0592 logPb2+Sn2+
Ecell= 0.010 + 0.0295 logPb2+Sn2+
when Ecell is negative then cell reaction reverses,
if Pb2+Sn2+ is in between 0.1 to 0.9 then, cell reaction reverse.