A silver oxide-zinc cell maintains a fairly constant voltage during discharge (1.50V). The button form of the cell is used in hearing aids. The half-reactions involved are : `Zn(s)+2OH^(-)(aq) to Zn(OH)_(2)(s)+2e^(-)` `Ag_(2)O(s)+H_(2)O(l)+2e^(-) to 2Ag(s)+2OH^(-)(aq)` Will change in `[OH^(-)]` affect `E_(cell)` ?

A silver oxide-zinc cell maintains a fairly constant voltage during di

1.	A silver oxide-zinc cell maintains a fairly constant voltage during discharge (1.50V). The button form of the cell is used in hearing aids. The half-reactions involved are : `Zn(s)+2OH^(-)(aq) to Zn(OH)_(2)(s)+2e^(-)` `Ag_(2)O(s)+H_(2)O(l)+2e^(-) to 2Ag(s)+2OH^(-)(aq)` Will change in `[OH^(-)]` affect `E_(cell)` ?
Answer» Let us add the two half cell reactions : `Zn(s) +2OH^(-)(aq) to Zn (OH)_(2)(s)+2e^(-)` `(Ag_(2)O(s)+H_(2)O(l)+2e^(-) to 2Ag(s)+2OH^(-)(aq))/(Zn(s)+Ag_(2)O(s)+H_(2)O(l) to Zn(OH)_(2)(s)+2Ag(s)` As the `OH^(-)` ions are not involved in the final equation, any change in [OH^(-)] will not affect `E_(cell)`.

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