Question

What would happen to the measured cell potentials if 30 mL solution was used in each half-cell instead of 25 mL

Answer 1

Answer:

The answer is "$\bold{\log \frac{[0] mole}{[R]mole}}$"

Explanation:

$E_{cell} =E_{cell}^{\circ} - \frac{0.0591}{n}= \log\frac{[0]}{[R]}$

In the above-given equation, we can see from $E_{ceu}$, of both oxidant $conc^n$as well as the reactant were connected. however, weight decreases oxidant and reduction component concentration only with volume and the both of the half cells by the very same factor  and each other suspend

$\to \log \frac{\frac{\text{oxidating moles}}{25 \ ml}}{\frac{\text{moles of reduction}}{25 ml}} \ \ = \ \ \log \frac{\frac{\text{oxidating moles}}{30 \ ml}}{\frac{\text{moles of reduction}}{30 ml}}$

$\to {\log \frac{[0] mole}{[R]mole}}$