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The standard reduction potential of zinc...

The standard reduction potential of zinc and silver at 298 K are `E_((Zn^(2+)) /(Zn))^0 = -0.76 V`, `E_((Ag^(+))/(Ag))^0= 0.80V` Which of the following reactions actually takes place in a cell reaction?

A

`Zn^(2+)(aq) + 2Ag^(+)(aq) rarr Zn(s) + 2Ag(s)`

B

`Zn(s) + 2Ag^+(aq)rarr Zn^(2+)(aq) + 2Ag(s)`

C

`Zn^(2+)(aq) + 2Ag(s) rarr 2Ag^+(aq) + Zn(s)`

D

`Zn(s) + 2Ag(s) rarr Zn^(2+)(aq) + 2Ag^+`

Text Solution

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The correct Answer is:
To determine which reaction takes place in a cell reaction involving zinc and silver, we need to analyze the standard reduction potentials provided and identify the oxidation and reduction processes. ### Step-by-Step Solution: 1. **Identify the Standard Reduction Potentials:** - For Zinc: \[ E^\circ_{(Zn^{2+}/Zn)} = -0.76 \, V \] - For Silver: \[ E^\circ_{(Ag^+/Ag)} = 0.80 \, V \] 2. **Determine Which Species Will Be Reduced and Oxidized:** - The species with a higher standard reduction potential will be reduced, while the species with a lower standard reduction potential will be oxidized. - Since \(E^\circ_{Ag^+/Ag} (0.80 \, V) > E^\circ_{(Zn^{2+}/Zn)} (-0.76 \, V)\), silver ions (\(Ag^+\)) will be reduced, and zinc (\(Zn\)) will be oxidized. 3. **Write the Half-Reactions:** - **Oxidation (Anode):** Zinc is oxidized: \[ Zn (s) \rightarrow Zn^{2+} (aq) + 2e^- \] - **Reduction (Cathode):** Silver ions are reduced: \[ Ag^+ (aq) + e^- \rightarrow Ag (s) \] 4. **Balance the Half-Reactions:** - The oxidation reaction produces 2 electrons, while the reduction reaction consumes only 1 electron. To balance the electrons, we multiply the reduction half-reaction by 2: \[ 2Ag^+ (aq) + 2e^- \rightarrow 2Ag (s) \] 5. **Combine the Balanced Half-Reactions:** - Now, we can add the two half-reactions together: \[ Zn (s) + 2Ag^+ (aq) \rightarrow Zn^{2+} (aq) + 2Ag (s) \] 6. **Final Cell Reaction:** - The overall cell reaction is: \[ Zn (s) + 2Ag^+ (aq) \rightarrow Zn^{2+} (aq) + 2Ag (s) \] 7. **Identify the Correct Option:** - From the options provided, we check which one matches the final cell reaction. The correct option is the one that represents this reaction.
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The standard oxidation potential of Zn and Ag in water at 20^(@) C" are: " Zn(s) rarr Azn^(2+) (aq) + 2e^(-) E^(o) = 0.76 V Ag(s) rarr Ag^(+)(aq) +e^(-) E^(o) = - 0.80 V Which one of the following reactions actually takes place?

Given that, Ni^(2+)//Ni=0.25 V, Cu^(2+)//Cu=0.34 V , Ag^(+)//Ag=0.80 V and Zn^(2+)//Zn=-0.76 V Which of the following reaction under standard condition will not take place in the specified direction ?

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Standard reduction electrode potential of Zn^(2+)//Zn is -0.76V . This means:

Consider the cell : Zn|Zn^(2+)(aq)(1.0M)||Cu^(2+)(aq)(1.0M)||Cu Thee standard reduction potentials are 0.350V for Cu^(2+)(aq)+2e^(-)rarrCu and -0.763V for Zn^(2+)(aq)+2e^(-) rarr Zn a. Write the cell reaction. b. Calculate the EMF of the cell. c. Is the reaction spontaneous or not ?

For Zn^(2+) //Zn, E^(@) =- 0.76 , for Ag^(+)//Ag, E^(@) = -0.799V . The correct statement is

(a) A cell is prepared by dipping a zinc rod in 1M zinc sulphate solution and a silver electrode in 1M silver nitrate solution. The standard electrode potential given : E^(@)Zn_(2+1//Zn) = -0.76V, E^(@)A_(g+//)A_(g) = +0.80V What is the effect of increase in concentration of Zn^(2+) " on the " E_(cell) ? (b) Write the products of electrolysis of aqueous solution of NaCI with platinum electrodes. (c) Calculate e.m.f. of the following cell at 298 K: "Ni(s)"//"Ni"^(2+)(0.01M)////"Cu"^(2+)(0.1M)//"Cu(s)" ["Given"E_(Ni2+//Ni)^(@) = -0.025 V E_(Cu2+//Cu)^(@) = +0.34V] Write the overall cell reaction.

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