Question

If the cell $E_{cell}^0$for a given reaction has a negative value then which of the following gives correct relationships for the values of $\Delta {G^0}\& {K_{eq}}$?
A.$\Delta {G^0} < 0;{K_{eq}} > 1$
B.$\Delta {G^0} < 0;{K_{eq}} < 1$
C.$\Delta {G^0} > 0;{K_{eq}} < 1$
D.$\Delta {G^0} > 0;{K_{eq}} > 1$

Answer 1

Hint : The electromotive force (EMF) of a galvanic cell is defined as the electrical potential which causes the flow of current from one electrode to another when no current is drawn from the external circuit of the cell. The EMF of a galvanic cell will be positive when there is a tendency for the electrons to flow from left to right in the external circuit. The cell in which oxidation takes place on the left-hand electrode and reduction at the right-hand electrode has a positive value of EMF.

Complete step by step solution :
Step 1
The cell that converts chemical energy into electrical energy is called a galvanic cell. When the cell operates, i.e. the cell reaction takes place, it does electrical work by transferring electrical charge through an external circuit. The electrical work done by the cell corresponds to a decrease in the free energy of the system/cell.
Step 2
There are two possibilities with free energy change:
When the free energy change $\Delta G$of the system is negative, the system does electrical work on the surroundings.
When the free energy change $\Delta G$ of the system is positive, no electrical work is done by the system, i.e. no electric current is generated.
Step 3
We know,
$\Delta G = - nF{E_{cell}}$
When the cell reaction takes place under standard states, we have
$\Delta {G^0} = - nF{E^0}_{cell}$
Step 4
The standard cell potential is related to the equilibrium constant as follows:
${E^0}_{cell} = \dfrac{{2.303RT}}{{nF}}{\log _{10}}{K_c}$
$\Delta {G^0} = - nF \times \dfrac{{2.303RT}}{{nF}}{\log _{10}}{K_c}$
$\Delta {G^0} = - 2.303RT{\log _{10}}{K_c}$
Hence, from the discussion above if $E_{cell}^0$for a given reaction has a negative value then ${G^0} > 0$ that is it has a positive value, and in turn ${K_{eq}} < 1$ has a negative value. So, option c) is the correct answer for the given question.

Note : When ${E^0}_{cell}$ is positive, the cell reaction is feasible and will proceed spontaneously, when ${E^0}_{cell}$ is negative, the cell reaction will not be feasible and when ${E^0}_{cell}$ is zero, the cell reaction attains equilibrium.