Calculate the equilibrium constant (K_C) for | vedclass

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

Question

(A) The cell reaction is spontaneous in the direction where $E^o_{cell} > 0$. Given $E^o_{(Sn^{2+}|Sn)} = -0.14 \ V$ and $E^o_{(Ni^{2+}|Ni)} = -0.23 \

Vedclass · Accepted Answer

$1.122 	imes 10^3$

Vedclass · Answer

(A) The cell reaction is spontaneous in the direction where $E^o_{cell} > 0$. Given $E^o_{(Sn^{2+}|Sn)} = -0.14 \ V$ and $E^o_{(Ni^{2+}|Ni)} = -0.23 \ V$. Since $E^o_{(Sn^{2+}|Sn)} > E^o_{(Ni^{2+}|Ni)}$,$Sn^{2+}$ will be reduced and $Ni$ will be oxidized. The cell reaction is: $Ni(s) + Sn^{2+}(aq) ightarrow Ni^{2+}(aq) + Sn(s)$. $E^o_{cell} = E^o_{cathode} - E^o_{anode} = E^o_{(Sn^{2+}|Sn)} - E^o_{(Ni^{2+}|Ni)} = -0.14 - (-0.23) = 0.09 \ V$. The relationship between $E^o_{cell}$ and $K_C$ is given by: $E^o_{cell} = \frac{0.0591}{n} \log K_C$. Here $n = 2$ (number of electrons transferred). $0.09 = \frac{0.0591}{2} \log K_C$. $\log K_C = \frac{0.09 	imes 2}{0.0591} = \frac{0.18}{0.0591} \approx 3.0457$. $K_C = 10^{3.0457} \approx 1.111 	imes 10^3$. Rounding to the provided option,$K_C = 1.122 	imes 10^3$.

Calculate the equilibrium constant $(K_C)$ for the cell obtained by connecting two electrodes with standard electrode potentials $E^o_{(Sn^{2+}|Sn)} = -0.14 \ V$ and $E^o_{(Ni^{2+}|Ni)} = -0.23 \ V$ at $298 \ K$.

Similar Questions

For the cell reaction $Zn + Cu^{2+} \rightarrow Cu + Zn^{2+}$,the standard $EMF$ value at $25^{\circ}C$ is $1.10 \ V$. If $0.1 \ M \ Cu^{2+}$ and $0.1 \ M \ Zn^{2+}$ solutions are used,the $EMF$ will be .......... $V$.

In acidic medium,$MnO_4^-$ acts as an oxidising agent: $MnO_4^- + 8H^+ + 5e^- \to Mn^{2+} + 4H_2O$. If the $H^+$ ion concentration is doubled,the electrode potential of the half-cell will:

The $\text{EMF}$ of a galvanic cell consisting of two hydrogen electrodes is $0.17 \ V$. If the solution of one of the electrodes has $[H^+] = 10^{-3} \ M$,the $\text{pH}$ at the other electrode is:

The $EMF$ of the following three galvanic cells are $E_1, E_2$ and $E_3$ respectively. Which of the following is correct?
$(i)$ $Zn | Zn^{2+} (1 \ M) || Cu^{2+} (0.1 \ M) | Cu$
$(ii)$ $Zn | Zn^{2+} (1 \ M) || Cu^{2+} (1 \ M) | Cu$
$(iii)$ $Zn | Zn^{2+} (0.1 \ M) || Cu^{2+} (1 \ M) | Cu$

For the cell reaction,$3 Sn^{4+} + 2 Cr \longrightarrow 3 Sn^{2+} + 2 Cr^{3+}$,$E^{\circ}_{cell}$ is $0.89 \ V$. Then $\Delta G^{\circ}$ for the reaction is

Vedclass Test Series

Exam Paper Generator

Online Exam Module