1876 United States presidential election in Iowa information
Main article: 1876 United States presidential election
1876 United States presidential election in Iowa
← 1872
November 7, 1876
1880 →
Turnout
24.57% 6.45 pp[1]
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
Elections in Iowa
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Question 1
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Amendment 1
Cedar Rapids
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v
t
e
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.[2]
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.[2]
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.
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Global Information
6.45 pp[1]
