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Unknown concentration, pH or standard potential from a measured cell EMF

This type of question, asked 4 times: 14 Jun 2022, 29 Jun 2022, 29 Jul 2022, 5 Apr 2026 S2. All 4 questions below, oldest first.

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  1. Question 1

    14 Jun 2022

    54
    The cell potential for the following cell Pt∣H2( g)∣H+(aq)∣∣Cu2+(0.01M)∣Cu(s)Pt\left| H_{2}(\text{ }g) \right|H^{+}(aq)||Cu^{2 +}(0.01M) \mid Cu(s) is 0.576 V0.576\text{ }V at 298 K298\text{ }K. The pHpH of the solution is (Nearest integer)

  2. Question 2

    29 Jun 2022

    144
    The cell potential for the given cell at 298 K298\text{ }K Pt ∣H2\mid H_{2} (g, 1 bar) ∣H+\mid H^{+}(aq) ∥Cu2+(aq)∣Cu(s)\parallel Cu^{2 +}(aq) \mid Cu(s) is 0.31 V0.31\text{ }V. The pHpH of the acidic solution is found to be 3 , whereas the concentration of Cu2+Cu^{2 +} is 10−xM10^{- x}M. The value of xx is (Given: ECu2+/CuΘ=0.34 VE_{Cu^{2 +}/Cu}^{\Theta}= 0.34\text{ }V and 2.303RTF=0.06 V\frac{2.303RT}{F}= 0.06\text{ }V )

  3. Question 3

    29 Jul 2022

    144
    For a cell, Cu(s)∣Cu2+(0.001 M)∥Ag+(0.01 M)∣Ag(s)Cu(s) \mid Cu^{2 +}(0.001\text{ }M) \parallel Ag^{+}(0.01\text{ }M) \mid Ag(s), the cell potential is found to be 0.43 V0.43\text{ }V at 298 K298\text{ }K. The magnitude of standard electrode potential for Cu2+/CuCu^{2 +}/Cu is ____ ×10−2 V\times 10^{- 2}\text{ }V. [Given: EAg+/Ag∘=0.80 V and 2.303RTF=0.06 V]\left\lbrack \text{Given: }E_{Ag^{+}/Ag}^{\circ} = 0.80\text{ }V\text{ and }\frac{2.303RT}{F} = 0.06\text{ }V \right\rbrack

  4. Question 4

    5 Apr 2026 S2

    32
    One half cell in a voltaic cell is constructed by dipping silver rod in AgNO3AgNO_{3} solution of unknown concentration, other half cell is Zn rod dipped in 1 molar solution of ZnSO4ZnSO_{4}. A voltage of 1.60 V is measured at 298 K for this cell. What is the concentration of Ag+Ag^{+}ions used in terms of log⁡x(x=[Ag+])\log x\left( x =\left\lbrack Ag^{+} \right\rbrack \right)? EZn2+/ZnΘ=−0.76 V,EAg+/AgΘ=+0.80 VE_{Zn^{2 +}/Zn}^{\Theta}= - 0.76\text{ }V,E_{Ag^{+}/Ag}^{\Theta}= + 0.80\text{ }V, 2.303RTF=0.059 V\frac{2.303RT}{F}= 0.059\text{ }V

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