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1888 United States presidential election in Iowa

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1888 United States presidential election in Iowa

← 1884 November 6, 1888 1892 →
 
Nominee Benjamin Harrison Grover Cleveland
Party Republican Democratic
Home state Indiana New York
Running mate Levi P. Morton Allen G. Thurman
Electoral vote 13 0
Popular vote 211,603 179,877
Percentage 52.36% 44.51%

President before election

Grover Cleveland
Democratic

Elected President

Benjamin Harrison
Republican

The 1888 United States presidential election in Iowa took place on November 6, 1888, as part of the 1888 United States presidential election. Voters chose 13 representatives, or electors to the Electoral College, who voted for president and vice president.

Iowa voted for the Republican nominee, Benjamin Harrison, over the Democratic nominee, incumbent President Grover Cleveland. Harrison won the state by a margin of 7.85%.

Results

1888 United States presidential election in Iowa[1]
Party Candidate Running mate Popular vote Electoral vote
Count % Count %
Republican Benjamin Harrison of Indiana Levi Parsons Morton of New York 211,603 52.36% 13 100.00%
Democratic Grover Cleveland of New York Allen Granberry Thurman of Ohio 179,877 44.51% 0 0.00%
Labor Alson Streeter of Illinois Charles E. Cunningham of Arkansas 9,105 2.25% 0 0.00%
Prohibition Clinton Bowen Fisk of New Jersey John Anderson Brooks of Missouri 3,550 0.88% 0 0.00%
Total 404,135 100.00% 13 100.00%


Notes

References

  1. ^ "1888 Presidential General Election Results - Iowa". U.S. Election Atlas. Retrieved 23 December 2013.