Question

The cell potential for the electrochemical reaction shown below depends upon the
$C{{l}^{-}} and C{{u}^{2+}}$ concentrations. Calculate the cell potential (in V) at . if $[C{{u}^{2+}}]=3.5M$ and $[C{{l}^{-}}]=1.7M$.
\[C{{u}^{2+}}(aq)+2C{{l}^{-}}(aq)+2Ag(s)Cu(s)+2AgCl(s);E=0.12V\]
(A) 0.15V
(B) -0.15V
(C) 0.30V
(D) -0.30V

Answer 1

Hint: The cell potential can be obtained by using Nernst equation, which relates the Gibbs free energy and cell potential in electrochemistry. It is very helpful in determining cell potential, equilibrium constant etc.

Complete step by step solution:
- Nernst equation is a general equation that relates the Gibbs free energy and cell potential in electrochemistry.
- It can be used to find the cell potential at any moment during a reaction or at conditions other than standard-state. It is mathematically represented as,
\[{{E}_{cell}}=E_{cell}^{0}-\dfrac{0.059}{n}\log \dfrac{1}{{{[C{{l}^{-}}]}^{2}}[C{{u}^{2+}}]}\]
- The voltage or electric potential difference across the terminals of a cell when no current is drawn from it. The electromagnetic force (emf) is the sum of the electric potential differences produced by a separation of charges (electrons or ions) that can occur at each interface in the cell.
- To calculate the cell potential, we will use the Nernst Equation.
- After substituting the values in the equation, we get,
\[{{E}_{cell}}=0.12-\dfrac{0.059}{2x0}\log \dfrac{1}{{{(1.7)}^{2}}(3.5)}\]
\[{{E}_{cell}}=0.12-\dfrac{0.059}{2x0}\log (0.099)\]
\[{{E}_{cell}}=0.12-(-0.029)\]
\[{{E}_{cell}}=0.12+0.029\]
\[{{E}_{cell}}=0.15V\]
- Therefore, the cell potential, \[{{E}_{cell}}=0.15V\].

The answer to the question is (A) 0.15V.

Note: The Nernst equation defines the relationship between cell potential to standard potential and to the activities of the electrically active (electroactive) species. It relates the effective concentrations (activities) of the components of a cell reaction to the standard cell potential.