The Nernst equation for the reaction, A2+ + 2e- → B, in terms of the free energy change is
A. ΔG = ΔG° + 2.303 RT log $$\frac{{\left[ B \right]}}{{\left[ A \right]}}$$
B. ΔG = ΔG° - 2.303 RT log $$\frac{{\left[ B \right]}}{{\left[ A \right]}}$$
C. -ΔG = -ΔG° + 2.303 RT log $$\frac{{\left[ B \right]}}{{\left[ A \right]}}$$
D. ΔG = -ΔG° + 2.303 RT log $$\frac{{\left[ B \right]}}{{\left[ A \right]}}$$
Answer: Option A
This Question Belongs to Engineering Chemistry >> Electrochemistry
A. $$2.303\frac{{RT}}{F}$$
B. $$ - 2.303\frac{{RT}}{F}$$
C. $$E_{{M^ + }|m}^ \circ + 2.303\frac{{RT}}{F}$$
D. $$E_{{M^ + }|m}^ \circ - 2.303\frac{{RT}}{F}$$
Given that E°(Fe3+, Fe) = -0.4 V and E°(Fe2+, Fe) = -0.44 V, the value of E°(Fe3+, Fe2+) is
A. 0.76V
B. -0.40 V
C. -0.76 V
D. 0.40 V
A. decrease by a factor of 2
B. increase by a factor of 2
C. decrease by a factor of 4
D. increase by a factor of 4
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