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\(^{2+}\).
Consider the following half cell reaction $ \text{Cr}_2\text{O}_7^{2-} (\text{aq}) + 6\text{e}^- + 14\text{H}^+ (\text{aq}) \longrightarrow 2\text{Cr}^{3+} (\text{aq}) + 7\text{H}_2\text{O}(1) $
The reaction was conducted with the ratio of $\frac{[\text{Cr}^{3+}]^2}{[\text{Cr}_2\text{O}_7^{2-}]} = 10^{-6}$
The pH value at which the EMF of the half cell will become zero is ____ (nearest integer value)
[Given : standard half cell reduction potential $\text{E}^\circ_{\text{Cr}_2\text{O}_7^{2-}, \text{H}^+/\text{Cr}^{3+}} = 1.33\text{V}, \quad \frac{2.303\text{RT}}{\text{F}} = 0.059\text{V}$
20 mL of sodium iodide solution gave 4.74 g silver iodide when treated with excess of silver nitrate solution. The molarity of the sodium iodide solution is _____ M. (Nearest Integer value) (Given : Na = 23, I = 127, Ag = 108, N = 14, O = 16 g mol$^{-1}$)