For questions involving \(H_2\) liberation:
• Compare the reduction potential of the metal ion with the standard hydrogen electrode (SHE).
• Negative reduction potentials indicate the ability to liberate \(H_2\) gas.
V2+ and Cr2+
V2+ and Mn2+
Cr2+ and Co2+
Mn2+ and Co2+
- Metal cations with negative values of reduction potential (\(\text{M}^{3+}/\text{M}^{2+}\)) or positive values of oxidation potential (\text{M}^{2+}/\text{M}^{3+}\)) can reduce H\(^+\) ions and liberate H\(_2\) gas from dilute acid.
- For the given metals:
V\(^{2+}\) has a reduction potential of \(-0.26~\text{V}\).
Cr\(^{2+}\) has a reduction potential of \(-0.41~\text{V}\).
- Both values are negative, meaning V\(^{2+}\) and Cr\(^{2+}\) can reduce H\(^+\) ions to liberate H\(_2\) gas.
Final Answer: \((3)\) V\(^{2+}\) and Cr\(^{2+}\).
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