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# Calculate cell potential at ${\text{298}}\,{\text{K}}$ for following galvanic cell$Cd\left( s \right)|C{d^{2 + }}\left( {aq} \right)\left( {0.1\,M} \right)\,||{H^ + }\left( {aq} \right)\left( {0.1\,M} \right)|{H_2}\left( g \right)\left( {0.5\,atm} \right)|Pt$$E_{c{d^{2 + }}/cd}^ + = \, - 0.40\,V$A. 0.38B.-0.38C.0.36D.0.40

Last updated date: 09th Sep 2024
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Hint: In an electrolytic cell, there are two electrodes. These two electrodes are cathode and anode. At the cathode, reduction occurs and at the anode oxidation occurs.

We know that in a galvanic cell the electrode potential is given as,
${E^0} = E_{cathode}^0 - E_{anode}^0$
Here,
${E^0}$= electrode potential of the cell
$E_{cathode}^0$= electrode potential of the cathode
$E_{anode}^0$= electrode potential of the anode
It is given,
$E_{c{d^{2 + }}/cd}^ + = \, - 0.40\,V$
$\left[ {C{d^{2 + }}} \right] = \,0.1\,{\text{M}}$
$\left[ {{H^{2 + }}} \right] = \,0.5\,{\text{M}}$
We will now calculate the value of ${E^0}$.
${E^0} = E_{cathode}^0 - E_{anode}^0 \\ = \,0 - \,\left( { - 0.4} \right) \\ = \,0.4 \\$
Using the given values, we will now calculate the value of $E$ using the Nernst equation,
$E = {E^0} - \dfrac{{0.0592\,V}}{n}{\log _{10}}Q$ ……(1)
Here,
$Q$= reaction quotient
$n$= number of electrode transfer
${E^0}$= electrode potential
Now, we will put the given values in equation (1)
$E = {E^0} - \dfrac{{0.059}}{2}\log \dfrac{{\left[ {c{d^{2 + }}} \right]\left[ H \right]}}{{{{\left[ {{H^ + }} \right]}^2}}} \\ = 0.4 - \dfrac{{0.059}}{2}\log \dfrac{{\left( {0.1} \right) \times \left( {0.5} \right)}}{{{{\left( {0.1} \right)}^2}}} \\ = 0.4 - 0.0295 \times \log \dfrac{{5 \times {{10}^{ - 2}}}}{{{{10}^{ - 2}}}} \\ = 0.379 \\ \simeq 0.38 \\$
Therefore, cell potential at ${\text{298}}\,{\text{K}}$ for the following galvanic cell is 0.38.

So, the correct answer is Option A.

Note:
An electrochemical cell produces electrical energy. This cell can convert the chemical energy into the electrical energy. An electrochemical cell consists of two half cells. In the electrochemical cell the two electrodes, the cathode and the anode are connected through a salt bridge. The reduction takes place at the cathode and the oxidation takes place at the anode. A voltmeter is also connected joining the cathode and the anode. A cathode is represented by a positive sign as the electrons go into the cathode and an anode is represented by a negative sign as the electrons come off the anode in the cell.
There are two common electrochemical cells. These are galvanic cells and electrolytic cells. In the galvanic cell, chemical energy is converted to electrical energy and in the electrolytic cell, electrical energy is converted into chemical energy.