The standard electrode potential for V, Cr, Mn and Co are - 0.26 V, - 0.41 V, + 1.57 V and + 1.97 V, respectively. The metal ions which can liberate from a dilute acid are [2023]

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Q.
The standard electrode potential $(M^{3 +} / M^{2 +})$ for V, Cr, Mn and Co are - 0.26 V, - 0.41 V, + 1.57 V and + 1.97 V, respectively. The metal ions which can liberate $H_{2}$ from a dilute acid are [2023]

1 $V^{2 +}$ and ${Mn}^{2 +}$

2 ${Mn}^{2 +}$ and ${Co}^{2 +}$

3 ${Cr}^{2 +}$ and ${Co}^{2 +}$

4 $V^{2 +}$ and ${Cr}^{2 +}$

Ans.
(4)

The metal ion for which have less value of reduction potential that $(E_{H^{+} / H_{2}}^{\circ} = 0)$ can release $H_{2}$ on reaction with dilute acid. So, $V^{+ 2}$ and ${Cr}^{+ 2}$ reduce $H^{+}$ to $H_{2}$ .

Solution

Q.
The standard electrode potential $(M^{3 +} / M^{2 +})$ for V, Cr, Mn and Co are - 0.26 V, - 0.41 V, + 1.57 V and + 1.97 V, respectively. The metal ions which can liberate $H_{2}$ from a dilute acid are [2023]

1 $V^{2 +}$ and ${Mn}^{2 +}$

2 ${Mn}^{2 +}$ and ${Co}^{2 +}$

3 ${Cr}^{2 +}$ and ${Co}^{2 +}$

4 $V^{2 +}$ and ${Cr}^{2 +}$

Ans.
(4)

The metal ion for which have less value of reduction potential that $(E_{H^{+} / H_{2}}^{\circ} = 0)$ can release $H_{2}$ on reaction with dilute acid. So, $V^{+ 2}$ and ${Cr}^{+ 2}$ reduce $H^{+}$ to $H_{2}$ .

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Solution

Q. The standard electrode potential (M3+/M2+) for V, Cr, Mn and Co are - 0.26 V, - 0.41 V, + 1.57 V and + 1.97 V, respectively. The metal ions which can liberate H2 from a dilute acid are [2023]

1 V2+ and Mn2+

2 Mn2+ and Co2+

3 Cr2+ and Co2+

4 V2+ and Cr2+

Ans. (4) The metal ion for which have less value of reduction potential that (EH+/H2∘=0) can release H2 on reaction with dilute acid. So, V+2 and Cr+2 reduce H+ to H2.

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Q.
The standard electrode potential $(M^{3 +} / M^{2 +})$ for V, Cr, Mn and Co are - 0.26 V, - 0.41 V, + 1.57 V and + 1.97 V, respectively. The metal ions which can liberate $H_{2}$ from a dilute acid are [2023]

1 $V^{2 +}$ and ${Mn}^{2 +}$

2 ${Mn}^{2 +}$ and ${Co}^{2 +}$

3 ${Cr}^{2 +}$ and ${Co}^{2 +}$

4 $V^{2 +}$ and ${Cr}^{2 +}$

Ans.
(4)

The metal ion for which have less value of reduction potential that $(E_{H^{+} / H_{2}}^{\circ} = 0)$ can release $H_{2}$ on reaction with dilute acid. So, $V^{+ 2}$ and ${Cr}^{+ 2}$ reduce $H^{+}$ to $H_{2}$ .