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HOME CBSE 12 CHEMISTRY SOLUTIONS
✓ 3 MARKS Numericals
🔥 APPEARED IN 3 BOARD SETS
Calculate emf of the following cell at $298\text{ K}$ : $\text{Sn} \mid \text{Sn}^{2+} (0.001\text{ M}) \parallel \text{H}^+ (0.01\text{ M}) \mid \text{H}_{2\text{(g)}} (1\text{ bar}) \mid \text{Pt}_{\text{(s)}}$ Given : $E^\circ_{\text{Sn}^{2+}/\text{Sn}} = - 0.14\text{ V}$, $E^\circ_{\text{H}^+/\text{H}_2} = 0.00\text{ V} [\log 10 = 1]$
YearPaper CodeSetPosition
CBSE 2026Series 56-2Set 1Question 23
CBSE 2026Series 56-2Set 2Question 23
CBSE 2026Series 56-2Set 3Question 23
  • Step 1: Stating the fundamental formula / law statement (+1 Mark)
  • Step 2: Intermediate mathematical substitution or proof steps (+1 Mark)
  • Step 3: Correct final value or conclusion with mandatory SI units (+1 Mark)
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