1888 United States presidential election in Kentucky
The 1888 United States presidential election in Kentucky took place on November 6, 1888. All contemporary thirty-eight states were part of the 1888 United States presidential election. Kentucky voters chose thirteen electors to the Electoral College, which selected the president and vice president.
Ever since the Civil War, Kentucky had been shaped politically by divisions created by that war between secessionist, Democratic counties and Unionist, Republican ones, although the state as a whole leaned Democratic throughout this era and the GOP would never carry the state during the Third Party System at either presidential or gubernatorial level. What would become a long-lived partisan system after the state was freed from the direct control of former Confederates would not be seriously affected by the first post-war insurgency movement – that of the Greenback Party at the tail end of the 1870s in the secessionist Jackson Purchase region. Incumbent president Grover Cleveland lost four points on his 1884 performance, but still carried the state comfortably against GOP nominee Benjamin Harrison.
Image: Stephen Grover Cleveland
Image: Benjamin Harrison 1896
1888 United States presidential election
The 1888 United States presidential election was the 26th quadrennial presidential election, held on Tuesday, November 6, 1888. Republican nominee Benjamin Harrison, a former U.S. senator from Indiana, defeated incumbent Democratic President Grover Cleveland of New York. It was the third of five U.S. presidential elections in which the winner did not win the national popular vote, which would not occur again until the 2000 US presidential election.
Image: Benjamin Harrison 1896
Image: Stephen Grover Cleveland
Man leaning on Harrison and Morton campaign ball.
President Grover Cleveland