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Electrochemistry: JEE Mains 2024 1st Feb Shift 1

Medium· Nernst equation· +4 / 0
We are given with following cell reaction : 2H++2e−⟶H22H^++2e^-\longrightarrow H_2 PH2=2P_{H_2}=2 atm [H+]=1[H^+]=1 M (2.303RTF=0.06)\left(\frac{2.303RT}{F}=0.06\right) If EcellE_{cell} for reaction is given by −x×10−3-x\times10^{-3} V, find out x.
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  • Previous-year question from JEE Mains 2024 1st Feb Shift 1.
  • Tagged medium difficulty.

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