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JEE Main 2023 (Online) 29th January Evening Shift
Numerical
+4
-1
Change Language

The equilibrium constant for the reaction

$$\mathrm{Zn(s)+Sn^{2+}(aq)}$$ $$\rightleftharpoons$$ $$\mathrm{Zn^{2+}(aq)+Sn(s)}$$ is $$1\times10^{20}$$ at 298 K. The magnitude of standard electrode potential of $$\mathrm{Sn/Sn^{2+}}$$ if $$\mathrm{E_{Z{n^{2 + }}/Zn}^\Theta = - 0.76~V}$$ is __________ $$\times 10^{-2}$$ V. (Nearest integer)

Given : $$\mathrm{\frac{2.303RT}{F}=0.059~V}$$

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2
JEE Main 2023 (Online) 29th January Morning Shift
Numerical
+4
-1
Change Language

Following figure shows dependence of molar conductance of two electrolytes on concentration. $$\Lambda \mathop m\limits^o $$ is the limiting molar conductivity.

JEE Main 2023 (Online) 29th January Morning Shift Chemistry - Electrochemistry Question 50 English

The number of $$\mathrm{\underline {incorrect} }$$ statement(s) from the following is ___________

(A) $$\Lambda \mathop m\limits^o $$ for electrolyte A is obtained by extrapolation

(B) For electrolyte B, $$\Lambda \mathop m\limits $$ vs $$\sqrt c$$ graph is a straight line with intercept equal to $$\Lambda \mathop m\limits^o $$

(C) At infinite dilution, the value of degree of dissociation approaches zero for electrolyte B.

(D) $$\Lambda \mathop m\limits^o $$ for any electrolyte A and B can be calculated using $$\lambda^\circ$$ for individual ions

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3
JEE Main 2023 (Online) 25th January Evening Shift
Numerical
+4
-1
Change Language

$$Pt(s)|{H_2}(g)(1\,bar)|{H^ + }(aq)(1\,M)||{M^{3 + }}(aq),{M^ + }(aq)|Pt(s)$$

The $$\mathrm{E_{cell}}$$ for the given cell is 0.1115 V at 298 K when $${{\left[ {{M^ + }(aq)} \right]} \over {\left[ {{M^{3 + }}(aq)} \right]}} = {10^a}$$

The value of $$a$$ is ____________

Given : $$\mathrm{E_{{M^{3 + }}/{M^ + }}^\theta = 0.2}$$ V

$${{2.303RT} \over F} = 0.059V$$

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4
JEE Main 2023 (Online) 25th January Morning Shift
Numerical
+4
-1
Change Language

Consider the cell

$$\mathrm{Pt(s)|{H_2}(g)\,(1\,atm)|{H^ + }\,(aq,[{H^ + }] = 1)||F{e^{3 + }}(aq),F{e^{2 + }}(aq)|Pt(s)}$$

Given $$\mathrm{E_{F{e^{3 + }}/F{e^{2 + }}}^o = 0.771\,V}$$ and $$\mathrm{E_{{H^ + }/1/2\,{H_2}}^o = 0\,V,\,T = 298\,K}$$

If the potential of the cell is 0.712 V, the ratio of concentration of Fe$$^{2+}$$ to Fe$$^{3+}$$ is _____________ (Nearest integer)

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