Consider the following electrochemical cell at 298K 2 2 Pt HSnO aq Sn(OH) (aq) Bi O s Bi s . If the — JEE Mains Chemistry Past Papers Chemistry Question
Question
Consider the following electrochemical cell at 298K 2 2 Pt HSnO aq Sn(OH) (aq) Bi O s Bi s . If the reaction quotient at a given time is 10 6, then the cell EMF (Ecell) is _______× 10 –1V (Nearest integer). Given the standard half-cell reduction potential as – 3 Bi O /Bi,OH E 0.44V and 6 0 Sn(OH) /HSnO ,OH E 0.90V
Answer: .
💡 Solution & Explanation
cell E 0.44 ( 0.90) = + 0.46 V Applying Nernst equation :- cell cell 0.06 E E logQ n cell 0.06 E 0.46 log10 Ecell = 4 × 10 –1 x = 4
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