Calculate the potential for the 1/2 cell below. కం 0.500m E5.00 -65atg ఇ +3Hం+ !e°E +0HT...
8. The standard cell potential (E°cell) for the reaction below is +1.10 V. Calculate the cell potential for this reaction when the concentration of [Cu2+] = 1.0 × 10-5 M and [Zn2+] = 3.5 M. Zn (s) + Cu2+ (aq) → Cu (s) + Zn2+ (aq)
Calculate the cell potential (E cell) for the following reaction that takes place in an electrochemical cell at 25 ^C. 2 ClO2 (g) + Zn (s) ---> 2 ClO2 (aq) + Zn^2+ (aq) [Zn^2+]=0.46 M, P-ClO2=0.015 atm, [ClO2]=0.75 M ClO2 (g) + e- ----> ClO2^- (aq) E^o= 0.95 V
For the reactions below, calculate the cell potential then determine which will be a voltaic cell and which will be an electrolytic cell. Reaction 1: Zn(OH)2(aq)+ 2Hg(l)+2Cl-(aq)> Zn(s) + Hg2Cl2(s)+2 OH-(aq) Reaction 2: Zn(s) + Hg2Cl2)s) + 2 OH-(aq)> Zn(OH)2(aq)+ 2Hg(l) + 2Cl- (aq)
Use the standard half-cell potentials listed below to calculate the standard cell potential for the following reaction occurring in an electrochemical cell at 25�C. (The equation is balanced.) Sn(s) + 2 Ag+(aq) ? Sn2+(aq) + 2 Ag(s) Sn2+(aq) + 2 e- ? Sn(s) E� = -0.14 V Ag+(aq) + e- ? Ag(s) E� = +0.80 V a)-1.08 V b)+1.74 V c)-1.74 V d)+0.94 V e)+1.08 V
1) Calculate the cell potential, E°cell, for the following electrochemical reaction. Cu(s) + I2(g) → Cu+2(aq) + 2I-1(aq) E = ? a.-0.87 V b.-0.19 V c.0.19 V d.0.87 V 2) Calculate the equilibrium constant (Keq) for the following reaction at 25 °C? (Faraday's Constant = 96,500 J / V · mol) & (R = 8.314 J / mol · K) 2Fe+3(aq) + Sn+2(aq) → 2Fe+2(aq) + Sn+4(aq) K = ? a. 4.04 x 107 b. 3.44 x 1018 c. 4.48...
Calculate the cell potential difference, Eºcell (Ecell = Eºcathode - Eanode), given the standard reduction potential below, Cut2 (aq) + 2 e. ----> Cu (s) ° = 0.337 volts Pb+2 (aq) + 2 e--...-> Pb(s) ° = -0.126 volts 0.211 V 0.463 V -0.463 V -0.211 V In Part III, to light up a white LED bulb will require forward voltage (VF) at 20 mA. • 5 1.2. 2.0 Calculate the cell potential difference, Eºcell ( cell = Eºcathode -...
24) Use the standard half-cell potentials listed below to calculate the standard cell potential for the following reaction occurring in an electrochemical cell at 25°C. (The equation is balanced.) 26). 3 Cl2(8) +2 Fe(s) - 6 Cl(aq) + 2 Fe3+ (aq) 12(8)+2 e 2Cl(aq) E° 1.36 V 3C1tbe ! Fe3+(aq) + 3 e → Fe(s) 80036V 2Fe > 25€ Ep 3 + the A) 240 V B) 4.16V ) -1.32 V D) -1.40 V E) 1.32 V
Using the Nernst equation [E=E degree cell- (-.0592/n) log Q], calculate the cell potential if [Cr3+] = 0.10M and [Cu2+]= 0.0010M. E degree cell = -1.08V
Use the standard half-cell potentials listed below to calculate the standard cell potential and standard free energy Gº) for the following reaction occurring in an electrochemical cell at 25 °C. (The equation is balanced.) k(aq) + e-K(5) E* --2.93 V 12(s) + 2 0 - 2 (aq) E*-+0.54V a. +6.40 V & 670 KJ b. +1.85 V & 487K] C. 3.47 V & 241 Kb d. 3.47 V8 - 6709 e.+5,32 V &-+670K)
Calculate the electrostatic potential of the cell at below 25 ° C .Pt (p) | H 2 (g, 200 kPa) | HNO 3 (aq, 0.10 M) || AgNO3 (aq, 0.0100 M) | Ag (s) E ° for Ag + + e- ® Ag (s) is 0.80 V.