at The E, of the 2H+ + 2 e- → H2 half reaction is arbitrarily set...
If the potential of a hydrogen electrode based on the half reaction 2H+(aq)+2e−→H2(g) is 0.00V at pH 0.00, what is the potential of the same electrode at pH = 4.00?
4. The following balanced redox reaction occurs in voltaic cell at 25°C H2 (g) + 2Ag+ (aq) → 2H* (aq) + 2 Ag (s) 2H+ + 2e → H2 (g) EⓇ = 0.00V Ag (aq) + le → Ag(s) E = +0.80V a. Write the two half-reactions (oxidation and reduction) occurring in the cell. Clearly indicate which reaction shows oxidation and which shows reduction. Clearly indicate which reaction occurs at the anode and which reaction occurs at the cathode. (4...
1. Consider the following 2 half reactions Fumarate +2H + 2e- --> succinate E= 0.031 Pyruvate +2H + 2e- --> lactate E= -0.185 a. In each reaction identify the oxidized form and the reduced form. b. Write the overall reaction c. Identify what is getting oxidized and what is getting reduced. d. Calculate deltaE e. Calculate deltaE under the following conditions; fumarate= 0.18mM, succinate= 0.32 mM, pyruvate= 0.25mM, lactate= 0.28nM at 37 degrees.
Consider the following half-reactions: Half-reaction F2(g) + 2e - 2H+ (aq) + 2 Mn2+ (aq) + 2e E° (V) 2F (aq) 2.870V H2(g) 0.000V Mn(s) -1.180V The strongest oxidizing agent is: enter formula / / The weakest oxidizing agent is: / The weakest reducing agent is: The strongest reducing agent is: Will Mn2+ (aq) reduce F2(g) to F"(aq)? Which species can be reduced by H2(g)? If none, leave box blank. Consider the following half-reactions: Half-reaction E° (V) Br2(1) + 2e...
Pb2+(aq) + 2e− ⇌ Pb(s) E° = -0.126 V 2H+(aq) + 2e− ⇌ H2(g) E° = 0.000 V E°cell (in V)= 0.126 V 2. The electrochemical cell is comprised of a Pb electrode in a 1.67 × 100 M solution of Pb2+ (aq) coupled to a Pt electrode in a solution containing H+ (aq) where the pH of the solution is 0.37 and the partial pressure of H2(g) is 0.571 atm. The temperature of the cell is held constant at...
Consider the following half-reactions: Half-reaction E° (V) F2(g) +2e - →2F (aq) 2.870V 2H*(aq) + 2e - H2(g) 0.000V Cr3+ (aq) + 3e — Cr(s) -0.740V (1) The strongest oxidizing agent is: enter formula (2) The weakest oxidizing agent is: (3) The weakest reducing agent is: (4) The strongest reducing agent is: (5) Will F2(g) oxidize Cr(s) to Cr3+ (aq)? (6) Which species can be oxidized by H(aq)? If none, leave box blank.
Use standard reduction potentials to calculate the equilibrium constant for the reaction: 2Cu2 (aq)-H2(g2Cu (aq) 2H (aq) Hint: Carry at least 5 significant figures during intermediate calculations to avoid round off error when taking the antilogarithm Equilibrium constant than zero. AG for this reaction would be 8 more group attempts remaining Retry Entire Group Submit Answer Use standard reduction potentials to calculate the equilibrium constant for the reaction: 2H (aq)+Hg) H2() + Hg2 " (aq ) Hint: Carry at least...
Consider the following half-reactions: Half-reaction E° (V) 12(s) + 2e - →21"(aq) 0.535V 2H+ (aq) + 2e - → H2(g) 0.000V Cr3+(aq) + 3e —— Cr(s) -0.740V The strongest oxidizing agent is: enter formula The weakest oxidizing agent is: The weakest reducing agent is: The strongest reducing agent is: Will 12(s) reduce Cr3+(aq) to Cr(s)? — Which species can be reduced by H2(g)? If none, leave box blank. Use the References to access important values if needed for this question....
39. Given: E +0.46 V Eo- +0.34 V Cu(s) + 2 Ag+ → 2 Ag (s) + Cu2. Cu2+ + H2 (g) → Cu (s) + 2 H' Find the standard potential for the cell reaction for 2 Ag + H2 (g)-2 Ag+2H +0.80 V b. a. +0.40 V +0.12 V d. c. -0.12 v none of these e. 40. Given: Fe (s) + 2 Ag' (aq) Fe (ag)+ 2 Ag (s) with E ell -1.24 V What is the...
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(4) Consider a cell at 298 K in which the reaction is Pb (s)+2H (aa-Pb2+ (ag) +H2 (g) (a) Cr ions are added to the Pb|Pb2+ half-cell to precipitate PbCh. The cell voltage is then measured to be +0.210 V. If [H]-1.0 M and the partial pressure of hydrogen is 1.0 atm, what is [Pb2*]? (answer: 1.57 x 103 M) (b) Calculate the Kp of PbCh if [Cl]-0.10 M in the PblPb2 half-cell. (answer: 1.57 x...