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निम्नलिखित सेल के विधुत वाहक बल (emf) की गणना कीजिए -
`Cu+Cu^(2+)(1M)||Ag^(+)(1M)|Ag`
`E_(Cu^(2+)//Cu)^(@)=+0.34` वोल्ट, `E_(Ag^(+)//Ag)^(@)=+0.80` वोल्ट

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`becauseE_(R.P._(Cu))^(@)=+0.34V:.E_(O.P._(Cu))^(@)=-0.34V`
`becauseE_(R.P._(Ag))^(@)=+0.80V:.E_(O.P._(Ag))^(@)=+0.80V`
चूँकि Cu का `E_(O.P.)^(@)` अधिक है अतः Cu ऐनोड व Ag कैथोड सेल अभिक्रिया निम्न प्रकार होगी
`{:(" "CutoCu^(2+)+2e),(ul(2Ag^(+)+2eto2Ag" ")),(Cu+2Ag^(+)toCu^(2+)+2Ag):}`
सेल का `"emf"(E_("cell")^(@))=E_(O.P._(Cu))^(@)+E_(R.P.Ag)^(@)=-0.34+0.80=0.46V`

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