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Electrochemistry question

2002 · Shift 0 · Q33
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Electrochemistry question

2002 · Shift 0 · Q33

JEE MainChemistryElectrochemistryMCQ+4 / −1
EMF of a cell in terms of reduction potential of its left and right electrodes is :
  1. A
    E = Eleft - Eright
  2. B
    E = Eleft + Eright
  3. C
    E = Eright - Eleft
  4. D
    E = -(Eright + Eleft)
View written solutionFree

Correct answer: C

  1. For a galvanic cell written as Left electrode  ∣  solution  ∣∣  solution  ∣  Right electrode\text{Left electrode} \; | \; \text{solution} \; || \; \text{solution} \; | \; \text{Right electrode}Left electrode∣solution∣∣solution∣Right electrode if both electrode potentials are given as reduction potentials, then the cell emf is: Ecell=Ecathode−EanodeE_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}}Ecell​=Ecathode​−Eanode​

  2. In standard cell notation, the right electrode is the cathode and the left electrode is the anode. So, Ecell=Eright−EleftE_{\text{cell}} = E_{\text{right}} - E_{\text{left}}Ecell​=Eright​−Eleft​

  3. Now compare with the given options:

  • A: E=Eleft−ErightE = E_{\text{left}} - E_{\text{right}}E=Eleft​−Eright​ ❌
  • B: E=Eleft+ErightE = E_{\text{left}} + E_{\text{right}}E=Eleft​+Eright​ ❌
  • C: E=Eright−EleftE = E_{\text{right}} - E_{\text{left}}E=Eright​−Eleft​ ✅
  • D: E=−(Eright+Eleft)E = -(E_{\text{right}} + E_{\text{left}})E=−(Eright​+Eleft​) ❌

Therefore, the correct expression is: E=Eright−EleftE = E_{\text{right}} - E_{\text{left}}E=Eright​−Eleft​

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