Consider the following electrochemical cell a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The cell reaction is: $3HSnO_2^- + Bi_2O_3 + 3H_2O + 6OH^- \rightarrow 3Sn(OH)_6^{2-} + 2Bi$.The number of electrons involved in the balanced equa

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) The cell reaction is: $3HSnO_2^- + Bi_2O_3 + 3H_2O + 6OH^- ightarrow 3Sn(OH)_6^{2-} + 2Bi$.The number of electrons involved in the balanced equation is $n = 6$.The standard cell potential is $E^0_{	ext{cell}} = E^0_{	ext{cathode}} - E^0_{	ext{anode}} = -0.44 - (-0.90) = +0.46 \ V$.Applying the Nernst equation at $298 \ K$: $E_{	ext{cell}} = E^0_{	ext{cell}} - \frac{0.0591}{n} \log Q$.Using $0.06$ as an approximation for $\frac{0.0591}{1}$: $E_{	ext{cell}} = 0.46 - \frac{0.06}{6} \log(10^6)$.$E_{	ext{cell}} = 0.46 - 0.01 	imes 6 = 0.46 - 0.06 = 0.40 \ V$.Thus, $E_{	ext{cell}} = 4 	imes 10^{-1} \ V$.The value is $4$.

Similar Questions

For the cell $Zn | Zn^{2+}(0.01 \, M) || Fe^{2+}(0.001 \, M) | Fe$ at $25^o C$,the $E_{cell} = 0.2905 \, V$. The equilibrium constant $K_c$ is:

The standard electrode potential for the following reaction is $+1.33 \ V$. What is the potential at $pH = 2.0$ for the reaction: $Cr_2O_7^{2-} (aq, 1 \ M) + 14H^{+} (aq) + 6e^{-} \rightarrow 2Cr^{3+} (aq, 1 \ M) + 7H_2O (l)$?

The e.m.f. of the cell $Ag | Ag^{+}(0.1 \ M) || Ag^{+}(1 \ M) | Ag$ at $298 \ K$ is ........... $V$.

The $emf$ of cell $Tl | Tl^+_{(0.001M)} || Cu^{2+}_{(0.01M)} | Cu$ is $0.83 \ V$ at $298 \ K$. It could be increased by:

If the $E^{\circ}_{cell}$ for a given reaction has a negative value,which of the following gives the correct relationships for the values of $\Delta G^{\circ}$ and $K_{eq}$ ?

Vedclass Test Series

Exam Paper Generator

Online Exam Module