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)` ?

Explanation

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)`.