The potential of the cell for the reaction, `M(s)+2H^(+) (1M)rightarrow H_(2) (g) (1atm)+M^(2+) (0.1m)`' is 1.500 V. The standard reduction potential for `M^(2+)` / M(s) couple is :

Options

A. `0.1470V`
B. `1.470V`
C. `1.47V`
D. none of these

Answer

C

Explanation

The cell is `M|M^(2+)(0.1M)||H^(+)(1M)|H_(2)(1atm)`
`E_(cell)=E_(cell)^(@)-(0.0591)/(2)"log"[M^(2+)]`
`therefore1.50=E_(cell)^(@)-(0.0591)/(2)xxlog(0.1)`
`=E_(cell)^(@)+0.02955`
or `E_(cell)^(@)=1.50-0.02955=1.47045V`
But `E_(cell)^(@)=E_(H^(+)//(1)/(2)H_(2))^(@)-E_(M^(2+)//M)^(@)`
i.e., `1.47=0-E_(M^(2+)//M)^(@)`
or `E_(Mg^(2+)//M)^(@)=-1.47M`