🤓 Based on our data, we think this question is relevant for Professor Nishida's class at WSU.

**Modified Nernst equation**:

$\overline{){{\mathbf{E}}}_{{\mathbf{cell}}}{\mathbf{=}}{\mathbf{E}}{{\mathbf{\xb0}}}_{{\mathbf{cell}}}{\mathbf{-}}\mathbf{\left(}\frac{\mathbf{0}\mathbf{.}\mathbf{0592}}{\mathbf{n}}\mathbf{\right)}{\mathbf{log}}\frac{\left[\mathrm{anode}\right]}{\left[\mathrm{cathode}\right]}}$

Ag^{+ }+ e^{-}→ Ag E° = +0.80 V

H_{2}O_{2} + 2H^{+} + 2e^{-}→ 2H_{2}O E° = +1.78 V

In log function:

- if anode > cathode → log (anode)/(cathode) = negative value → Ecell is higher than E°cell
- if cathode > anode → log (anode)/(cathode) = positive value → Ecell is lower than E°cell

For the half reactions:

**↓ E°**→ oxidation → anode**↑ E°**→ reduction → cathode

A galvanic cell is based on the following half-reactions at 25°C:

Ag^{+ }+ e^{-} → Ag

H_{2}O_{2} + 2H^{+} + 2e^{-} → 2H_{2}O

Predict whether E_{cell} is larger or smaller than E°_{cell} for the following cases.

b. [Ag^{+}] = 2.0 M, [H_{2}O_{2}] = 1.0 M, [H^{+}] = 1.0 x 10 ^{-7} M