1876 United States presidential election in Iowa

Main article: 1876 United States presidential election

1876 United States presidential election in Iowa

November 7, 1876
Turnout	24.57% 6.45 pp
Nominee Rutherford B. Hayes Samuel J. Tilden Party Republican Democratic Home state Ohio New York Running mate William A. Wheeler Thomas A. Hendricks Electoral vote 11 0 Popular vote 171,326 112,121 Percentage 58.50% 38.28%
County Results Hayes   40-50%   50-60%   60-70%   70-80%   80-90%   90-100% Tilden   40-50%   50-60%   60-70%
President before election Ulysses S. Grant Republican Elected President Rutherford B. Hayes Republican

The 1876 United States presidential election in Iowa took place on November 7, 1876, as part of the 1876 United States presidential election. Voters chose 11 representatives, or electors, to the Electoral College, who voted for president and vice president.

Iowa was won by Rutherford B. Hayes, the governor of Ohio (R-Ohio), running with Representative William A. Wheeler, with 58.50% of the vote, against Samuel J. Tilden, the former governor of New York (D–New York), running with Thomas A. Hendricks, the governor of Indiana and future vice president, with 38.28% of the popular vote.

The Greenback Party chose industrialist Peter Cooper and former representative Samuel Fenton Cary, received 3.22% of the vote. Iowa gave the Greenbacks both their largest number of total votes and percentage and Iowa would later be the last bastion of the congressional Greenbacks who held a House seat until 1889.

Results

1876 United States presidential election in Iowa
Party		Candidate	Running mate	Popular vote		Electoral vote
				Count	%	Count	%
	Republican	Rutherford B. Hayes of Ohio	William A. Wheeler of New York	171,326	58.50%	11	100.00%
	Democratic	Samuel J. Tilden of New York	Thomas A. Hendricks of Indiana	112,121	38.28%	0	0.00%
	Greenback	Peter Cooper of New York	Samuel Fenton Cary of Ohio	9,431	3.22%	0	0.00%
Total				292,878	100.00%	11	100.00%

See also

• United States presidential elections in Iowa