Q.57. For a general redox reaction Anode Cathode Which of the following statement is incorrect? (1

Q: What is the correct answer for this exam Chemistry Q57?

The correct answer is D. Sol. Anode: $n_2[Red_1 \rightarrow Ox_1^{n_1+} + n_1e^-]$ Cathode: $n_1[Ox_2^{n_2+} + n_2e^- \rightarrow Red_2^{n_2-}]$ Overall reaction: $n_2 Red_1 + n_1Ox_2^{n_2+} \rightarrow n_2Ox_1^{n_1+} + n_1 Red_2^{n_2-}$. The electrons do not appear in overall reaction. Nernst Equation: $E = E^\circ - \frac{2.303 RT}{nF} \log Q$. The slope of the straight line passing through origin for $E-E^\circ$ vs $\l

Q: Q.57. For a general redox reaction Anode Cathode Which of the following statement is incorrect? (1) The overall reaction can be written as (2) The electrons do not appear in the overall reaction…

The correct answer is D. Sol. Anode: $n_2[Red_1 \rightarrow Ox_1^{n_1+} + n_1e^-]$ Cathode: $n_1[Ox_2^{n_2+} + n_2e^- \rightarrow Red_2^{n_2-}]$ Overall reaction: $n_2 Red_1 + n_1Ox_2^{n_2+} \rightarrow n_2Ox_1^{n_1+} + n_1 Red_2^{n_2-}$. The electrons do not appear in overall reaction. Nernst Equation: $E = E^\circ - \frac{2.303 RT}{nF} \log Q$. The slope of the straight line passing through origin for $E-E^\circ$ vs $\l

Q: What are the options for this this exam Chemistry question?

A. The overall reaction can be written as $n_2 Red_1 + n_1Ox_2^{n_2+} \rightarrow n_2Ox_1^{n_1+} + n_1 Red_2^{n_2-}$ · B. The electrons do not appear in the overall reaction because electrons produced at the anode are consumed at the cathode. · C. Here n is the number of electrons transferred in redox reaction. · D. If the reaction is carried out reversibly, the electrical work done is equal to the ratio of charge and potential difference through which charge is moved.

Question

Q.57. For a general redox reaction Anode  Cathode  Which of the following statement is incorrect? (1) The overall reaction can be written as  (2) The electrons do not appear in the overall reaction because electrons produced at the anode are consumed at the cathode.[Figure: Graph of E-E° vs log₁₀Q showing three lines with slopes proportional to 1/n for n=1, n=2, n=3.] (3) Here n is the number of electrons transferred in redox reaction. (4) If the reaction is carried out reversibly, the electrical work done is equal to the ratio of charge and potential difference through which charge is moved.

FirstInTest · Accepted Answer

Sol. Anode: $n_2[Red_1 ightarrow Ox_1^{n_1+} + n_1e^-]$ Cathode: $n_1[Ox_2^{n_2+} + n_2e^- ightarrow Red_2^{n_2-}]$ Overall reaction: $n_2 Red_1 + n_1Ox_2^{n_2+} ightarrow n_2Ox_1^{n_1+} + n_1 Red_2^{n_2-}$. The electrons do not appear in overall reaction. Nernst Equation: $E = E^\circ - \frac{2.303 RT}{nF} \log Q$. The slope of the straight line passing through origin for $E-E^\circ$ vs $\log_{10}Q$ is $-\frac{2.303 RT}{nF}$. Electrical work = Charge $	imes$ Potential difference. Statement (4) is incorrect because electrical work done is equal to the product of charge and potential difference, not the ratio.

Answer

The overall reaction can be written as $n_2 Red_1 + n_1Ox_2^{n_2+} \rightarrow n_2Ox_1^{n_1+} + n_1 Red_2^{n_2-}$

Answer

The electrons do not appear in the overall reaction because electrons produced at the anode are consumed at the cathode.

Answer

Here n is the number of electrons transferred in redox reaction.

Answer

If the reaction is carried out reversibly, the electrical work done is equal to the ratio of charge and potential difference through which charge is moved.

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