Answer :-
Firstly from given half reactions, Eocell is calculated then using Nerst 's equation, cell potential is calculated.
The answer is given in the image,


יי Review 1 Constants l Periodic Table An electrochemical cell is based on these two half-reactions...
An electrochemical cell is based on these two half-reactions: Ox: Sn(s)→Sn2 + (aq, 1.55 mol L−1)+ 2e−, E∘anode=−0.14(V) Red: ClO2(g,0.200 bar)+e−→ClO2-(aq, 1.75 mol L−1), E∘cathode=0.95(V) Part A Compute the cell potential at 25 ∘C. Express your answer to two decimal places and include the appropriate units.
An electrochemical cell is based on these two half-reactions: Ox: Pb(s) Pb?" (aq, 0.21 mol L ')+2 e Erode = -0.13 (V) Red: Mno. (aq, 1.40 mol L-)+4 H+ (aq, 1.5 mol L-') +3 e MnO2 (s) + 2 H2O (1) Ecathode = 1.68 (V) Part A Compute the cell potential at 25°C. Express your answer to two decimal places and include the appropriate units. Ecell = Value Units Submit Request Answer
An electrochemical cell is based on these two half-reactions: Ox: Pb (8) Pb²+ (aq, 0.21 mol L-')+2 e Enode -0.13 (V) Red: MnO, (aq, 1.40 mol L-1) + 4H+ (aq, 1.5 mol L-) + 3 e MnO, (s) + 2 H20 (1) Eestbode - 1.68 (V) Part A Compute the cell potential at 25°C. Express your answer to two decimal places and include the appropriate units. T • Ea ? HẢ • Value • v Ecou -
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Write the cell notation for an electrochemical cell consisting of an anode where Al () is oxidized to AP+ (aq) and a cathode where Pb2+ (aq) is reduced to Pb (s). Assume all aqueous solutions have a concentration of 1 mol/L. Write the cell notation for an electrochemical cell consisting of an anode where Sn (s) is oxidized to Sn2+ (aq) and a cathode where H(aq) is reduced to H2 (g) at a platinum electrode. Assume...
MISSED THIS? Read Section 20.6 (Page) An electrochemical cell is based on the following two half-reactions: Part A oxidation: Sn (s) +Sn2+ (aq, 1.80 M )+2e reduction: C1O2(g, 0.210 atm )+e- +C10] (aq, 1.70 M) Compute the cell potential at 25°C. Express the cell potential to three significant figures. ΤΕΙ ΑΣφ ? Ecell = .839 Submit Previous Answers Request Answer X Incorrect; Try Again; One attempt remaining Provide Feedback
An electrochemical cell is based on the following two half-reactions: Part A oxidation: Sn (s) +Sn2+ (aq, 1.70 M)+2e reduction: CIO2(g, 0.265 atm )+e+C10(aq, 2.00 M) You may want to reference (Pages 865 - 869) Section 19.6 while completing this problem. Compute the cell potential at 25°C. Express the cell potential to three significant figures. ME PO ΑΣΦ ? Ecco V Submit Request Answer
An electrochemical cell is based on the following two half-reactions: Ox: Pb(s)→Pb2+(aq, 0.26 M )+2e- Red: MnO-4(aq, 1.50 M )+4H+(aq, 2.7 M )+3e-→MnO2(s)+2H2O(l) Compute the cell potential at 25 ∘C.
An electrochemical cell is based on the following two half-reactions: Ox: Pb(s)→Pb2+(aq, 0.18 M )+2e− Red: MnO−4(aq, 1.65 M )+4H+(aq, 1.9 M )+3e−→ MnO2(s)+2H2O(l) Compute the cell potential at 25 ∘C.
An electrochemical cell is based on the following two half-reactions: Ox: Pb(s)→Pb2+(aq, 0.29 M )+2e− Red: MnO−4(aq, 1.30 M )+4H+(aq, 1.7 M )+3e−→ MnO2(s)+2H2O(l) show work
An electrochemical cell is based on the following two half-reactions:oxidation: Sn(s)→Sn2+(aq,1.60 M) +2e-reduction: ClO2(g, 0.130 atm )+e−→ClO2-(aq, 1.55 MM )Compute the cell potential at 25°C