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Calculate the emf of the following cell ...

Calculate the emf of the following cell at `298K`:
`Fe(s)|Fe^(2+)(0.001M)||H^+(1M)|H_2(g)(1"bar"),Pt(s)` (Given`E_("Cell")^@=+0.44V)`

Text Solution

AI Generated Solution

To calculate the emf of the given electrochemical cell at 298K, we will follow these steps: ### Step 1: Identify the half-reactions In the given cell: - At the anode (oxidation), iron is oxidized: \[ \text{Fe(s)} \rightarrow \text{Fe}^{2+} + 2e^- \] ...
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(a) Write the cell reaction and calculate the emf of the following cell at 298 K : Sn (s) | Sn^(2+) (0.004 M) | | H^(+) (0.020 M) | H^2 (g) (1 bar) | Pt (s) [Given : E_(Sn^(2+)//Sn)^(@) = - 0.14 V ] (b) Give reasons : (i) On the basis of E° values, O_(2) gas should be liberated at anode but it is Cl_(2) gas which is liberated in the electrolysis of aqueous NaCl. (ii) Conductivity of CH_3COOH decreases on dilution.

Knowledge Check

  • Calculate the e.m.f. of the following cell at 298K : Pt(s)|Br_(2)(l)|Br^(-)(0.010M)||H^(+)(0.030M)|H_(2)(g)(1"bar")|Pt(s) Given : E_((1)/(2)Br_(2)//Br^(-))^(@)=+1.08V .

    A
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    B
    `1.288V`
    C
    `0.128V`
    D
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  • Calculate the emf of the following cell at 298K: Mg(s)|Mg^(2+)(0.1M)||Cu^(2+)(1.0xx10^(-3)M)|Cu(s) [Given= E_(Cell)^(@)=2.71V ]

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