The potential of a hydrogen electrode in a solution can be calculated using the Nernst equation:
\(E = E^0 - \frac{0.059}{n} \log \frac{1}{[\text{H}^+]}\)
Given:
- \(E^0 = 0\) (for the standard hydrogen electrode)
- pH = 3, so \([\text{H}^+] = 10^{-3} \, \text{M}\)
Substitute into the equation:
\(E = 0 - 0.059 \times \log(10^3) = -0.059 \times 3 = -0.177 \, \text{V}\)
Thus, the electrode potential is:
\(E = -17.7 \times 10^{-2} \, \text{V}\)
The Correct answer is: 18
For the given cell: \[ {Fe}^{2+}(aq) + {Ag}^+(aq) \to {Fe}^{3+}(aq) + {Ag}(s) \] The standard cell potential of the above reaction is given. The standard reduction potentials are given as: \[ {Ag}^+ + e^- \to {Ag} \quad E^\circ = x \, {V} \] \[ {Fe}^{2+} + 2e^- \to {Fe} \quad E^\circ = y \, {V} \] \[ {Fe}^{3+} + 3e^- \to {Fe} \quad E^\circ = z \, {V} \] The correct answer is:
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32