Question

The expermental setup for a typical `Zn-Ni` galvanic cell as shown below in figure:
(a). Identify M and X and determine cell potential at `25^(@)C`
(b). If concentration of `M^(2+)` ion changes to 1.0 M during its usage, what would be the new cell voltage?
(c). Describe, what would happen to cell voltage if salt bridge was removed.

Answer 1

Correct Answer - (a). `MtoZn` and `XtoNi`
`E_(cell)=+0.569V`
(b). `E_(cell)=+0.451` V (note, equal volume of electrolytes have been considered in the two half-cells.)
(c). Cell voltage becomes zero.
(a). `M=Zn&X=Ni`
`E_(cell)^(@)=-0.25+0.76=0.51V`
`E_(cell)=-0.25-(0.059)/(2)log(([Zn^(2+)])/([Ni^(2+)]))=0.51-(0.059)/(2)log((0.01)/(1))=0.569V`
(b). `Zn^(2+)=1M`
initial conc. conc.after time. `undersetunderset(0.01)(1)(Ni^(2+))+Zn(s)hArrundersetunderset(1)(0.01)(Zn^(2+))(aq)+Ni(s)`

`E_(cell)=0.51-(0.059)/(2)log((1)/(0.01))=0.451"volt"`
(c). circuit is disconnected & cell voltage `=0`