PYQs Homework Notes Mocks Board Daily Homework JEE Mains Chemistry Question type Concentration ratio of a redox couple from the measured EMF This type of question, asked 2 times: 25 Jan 2023, 30 Jan 2023. All 3 questions below, oldest first.
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Question 1 25 Jan 2023
146 P t ( s ) ∣ H 2 ( g ) ( 1 b a r ) ∣ H + ( a q ) ( 1 M ) ∥ M 3 + ( a q ) , M + ( a q ) ∣ P t ( s ) Pt(s)\mid H_{2}(g)(1\ bar)\mid H^{+}(aq)(1M)\parallel M^{3 +}(aq),M^{+}(aq)\mid Pt(s) P t ( s ) ∣ H 2 ( g ) ( 1 ba r ) ∣ H + ( a q ) ( 1 M ) ∥ M 3 + ( a q ) , M + ( a q ) ∣ P t ( s ) The E cell E_{\text{cell~}} E cell for the given cell is 0.1115 V 0.1115\text{ }V 0.1115 V at 298 K 298\text{ }K 298 K when [ M + ( a q ) ] [ M 3 + ( a q ) ] = 10 a \frac{\left\lbrack M^{+}(aq) \right\rbrack}{\left\lbrack M^{3 +}(aq) \right\rbrack}=10^{a} [ M 3 + ( a q ) ] [ M + ( a q ) ] = 1 0 a The value of a a a is Given : E M 3 + / M + θ = 0.2 V 2.303 R T F = 0.059 V \begin{matrix} & \text{~Given :~}E^{\theta}_{M^{3 +}/M^{+}}= 0.2\text{ }V \\ & \frac{2.303RT}{F}= 0.059\text{ }V \end{matrix} Given : E M 3 + / M + θ = 0.2 V F 2.303 R T = 0.059 V
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Question 2 25 Jan 2023
58 Consider the cell
P t ( s ) ∣ H 2 ( g ) ( 1 a t m ) ∣ H + ( a q , [ H + ] = 1 ) ∥ F e 3 + ( a q ) , F e 2 + ( a q ) ∣ P t ( s ) Pt(s)\left| H_{2}(\text{ }g)(1\text{ }atm) \right|H^{+}\left( aq,\left\lbrack H^{+} \right\rbrack= 1 \right)\parallel Fe^{3 +}(aq),Fe^{2 +}(aq) \mid Pt(s) P t ( s ) ∣ H 2 ( g ) ( 1 a t m ) ∣ H + ( a q , [ H + ] = 1 ) ∥ F e 3 + ( a q ) , F e 2 + ( a q ) ∣ P t ( s ) Given E F e 3 + / F e 2 + o = 0.771 V E_{Fe^{3 +}/Fe^{2 +}}^{o}= 0.771\text{ }V E F e 3 + / F e 2 + o = 0.771 V and E H + / 1 2 H 2 0 = 0 V , T = 298 K E_{H^{+}/\frac{1}{2}H_{2}}^{0}= 0\ V,T = 298\text{ }K E H + / 2 1 H 2 0 = 0 V , T = 298 K If the potential of the cell is 0.712 V 0.712\text{ }V 0.712 V , the ratio of concentration of F e 2 + Fe^{2 +} F e 2 + to F e 3 + Fe^{3 +} F e 3 + is (Nearest integer)
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Question 3 30 Jan 2023
53 Consider the cell P t ( s ) ∣ H 2 ( g , 1 a t m ) ∣ H + ( a q , 1 M ) ∣ ∣ F e 3 + ( a q ) , F e 2 + ( a q ) ∣ P t ( s ) Pt_{(s)}\left| H_{2}(\text{ }g,1\text{ }atm) \right|H^{+}(aq,1M)||Fe^{3 +}(aq),Fe^{2 +}(aq) \mid Pt(s) P t ( s ) ∣ H 2 ( g , 1 a t m ) ∣ H + ( a q , 1 M ) ∣∣ F e 3 + ( a q ) , F e 2 + ( a q ) ∣ P t ( s ) When the potential of the cell is 0.712 V 0.712\text{ }V 0.712 V at 298 K 298\text{ }K 298 K , the ratio [ F e 2 + ] / [ F e 3 + ] \left\lbrack Fe^{2 +} \right\rbrack/\left\lbrack Fe^{3 +} \right\rbrack [ F e 2 + ] / [ F e 3 + ] is (Nearest integer) Given: F e 3 + + e − = F e 2 + , E θ F e 3 + , F e 2 + ∣ P t = 0.771 Fe^{3 +}+e^{-}=Fe^{2 +},E^{\theta}Fe^{3 +},Fe^{2 +}\mid Pt = 0.771 F e 3 + + e − = F e 2 + , E θ F e 3 + , F e 2 + ∣ P t = 0.771 2.303 R T F = 0.06 V \frac{2.303RT}{F}= 0.06\text{ }V F 2.303 R T = 0.06 V
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