Question

: E∘ for the half cell \[Z{n^{2 + }}\mid Zn\] is\[ - 0.76\]. e.m.f. of the cell
\[Zn\mid Z{n^{2 + }}\left( {1M} \right){\text{ }}//{\text{ }}{H^ + }\left( {1M} \right){\text{ }}/{\text{ }}{H_2}\]​at \[1atm\] is:
1)\[ - 0.76\]
2)\[ + 0.76\]
3)\[ - 0.38\]
4)\[ + 0.38\]

Answer 1

Hint: The standard state potential of a cell is the cell’s potential under the standard state conditions that can be approximated with the concentrations of \[1mole/L\] (i.e.\[1M\]) and a pressure of \[1atm\] at\[25^\circ C\]. The overall standard potential of the cell by adding the potentials of the half-cells.

Complete step-by-step answer: In order to estimate the standard cell potential for a specific reaction, you must follow certain steps as mentioned below:
Write down the cell’s oxidation as well as reduction half-reactions.
Check the reduction potential i.e. \[{E^o}_{reduction}\]in case of the reduction half-reaction from a table of reduction potentials
Check the reduction potential in case of the reverse of reduction half-reaction and reverse the sign in order to get the oxidation potential. In case of the oxidation half-reaction, \[{E^o}_{oxidation} = - {\text{ }}{E^o}_{reduction}\]
Summation of the potentials of the half-cells in order to obtain the overall standard cell potential.
 \[{E^o}_{cell}\; = {\text{ }}{E^o}_{reduction}\; + {\text{ }}{E^o}_{oxidation}\]
Thus, in the present case, this equation becomes:
\[{E^o}_{cell} = {E_{{H^ + }}}{/_{{H_2}}} - {E_{Z{n^{2 + }}}}{/_{Zn}}\]
\[ = 0 - \left( { - 0.76} \right) = + 0.76\]

Hence, the correct answer is Option B.

Note: The potential energy which drives the redox reactions to be involved in the electrochemical cells refers to the potential for the anode to get oxidized and the potential for the cathode to get reduced. In\[{E_0}\], the superscript "\[0\]" indicates that these potentials are considered to be correct only in cases when concentrations are \[1M\] and correspondingly, pressures are\[1bar\].